Class 12 NCERT Maths – Exercise 7.4 Solutions

Q1. Evaluate \displaystyle \int \frac{3x^2}{x^6 + 1} dx

Solution:
Let u = x^3, so du = 3x^2 dx \implies x^2 dx = \frac{du}{3}.

Substitute:
x^6 + 1 = (x^3)^2 + 1 = u^2 + 1

So,
I = \int \frac{3x^2}{x^6 + 1} dx

I = \int \frac{du}{u^2 + 1}

I = \tan^{-1} u + C

I = \tan^{-1} (x^3) + C

Final Answer: \boxed {\int \frac{3x^2}{x^6 + 1} dx = \tan^{-1} (x^3) + C}

Q2. Evaluate \displaystyle \int \frac{1}{\sqrt{1 + 4x^2}} dx

Solution:
Write 1 + 4x^2 = 1^2 + (2x)^2, which matches the formula:

\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C

Let t = 2x,
so dt = 2 dx \implies dx = \frac{dt}{2}.

So,
I = \int \frac{1}{\sqrt{1 + 4x^2}} dx

I = \int \frac{1}{\sqrt{1 + t^2}} \cdot \frac{dt}{2}

I = \frac{1}{2} \int \frac{dt}{\sqrt{1 + t^2}}

I= \frac{1}{2} \log \left| t + \sqrt{1 + t^2} \right| + C

Substitute back t = 2x:

Final Answer: \boxed{ \int \frac{1}{\sqrt{1 + 4x^2}} = \frac{1}{2} \log \left| 2x + \sqrt{1 + 4x^2} \right| + C}

Q3. Evaluate \displaystyle \int \frac{1}{\sqrt{(2-x)^2 + 1}} dx

Solution:
Let u = 2-x \implies du = -dx.

So,
I = \int \frac{1}{\sqrt{(2-x)^2 + 1}} dx

I = -\int \frac{1}{\sqrt{u^2 + 1}} du

I = -\log \left| u + \sqrt{u^2 + 1} \right| + C

Substitute back u = 2-x:
I= -\log \left| 2 - x + \sqrt{(2-x)^2 + 1} \right| + C

I= \log \left| \frac {1}{2 - x + \sqrt{(2-x)^2 + 1}} \right| + C

Final Answer: \boxed {\int \frac{1}{\sqrt{(2-x)^2 + 1}} dx = \log \left| \frac {1}{2 - x + \sqrt{(x^2 -4x + 5}} \right| + C}

Q4. Evaluate \displaystyle \int \frac{1}{\sqrt{9 - 25x^2}} dx

Solution:
Write 9 - 25x^2 = 3^2 - (5x)^2,

I = \int \frac{1}{\sqrt{9 - 25x^2}} dx

I = \int \frac{1}{\sqrt{3^1 - (5x)^2}} dx

which matches the formula:
\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C

Let u = 5x \implies x = \frac{u}{5},\ dx = \frac{du}{5}.

So,
I = \int \frac{1}{\sqrt{9 - 25x^2}} dx

I = \int \frac{1}{\sqrt{9 - u^2}} \cdot \frac{du}{5}

I = \frac{1}{5} \int \frac{1}{\sqrt{9 - u^2}} du

I = \frac{1}{5} \sin^{-1} \left( \frac{u}{3} \right) + C

Substitute back u = 5x:

I = \frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C

Final Answer: \boxed{ \int \frac{1}{\sqrt{9 - 25x^2}} dx = \frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C }

Q5. Evaluate \displaystyle \int \frac{3x}{1 + 2x^4} dx

Solution:
I = \int \frac{3x}{1 + 2x^4} dx

So, x^4 = (x^2)^2 = u^2 and 2x^4 = 2u^2:

Let u = x^2 \implies du = 2x dx \implies x dx = \frac{du}{2}.

Now substitute x dx = \frac{du}{2}:

I = 3 \int \frac{\frac{du}{2}}{1 + 2u^2}

I = \frac{3}{2} \int \frac{du}{1 + 2u^2}

Now, write 1 + 2u^2 = 1 + (\sqrt{2} u)^2,
so:
I = \frac{3}{2} \int \frac{du}{1 + (\sqrt{2} u)^2}

I = \frac{3}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} u) + C

Substituting back ‘u’
I =\frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2} x^2) + C

Final Answer: \boxed{ \int \frac{3x}{1 + 2x^4} dx =\frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2} x^2) + C}

Q6. Evaluate \displaystyle \int \frac{x^2}{1 - x^6}; dx

Solution:
Let u = x^3 \implies du = 3x^2 dx \implies x^2 dx = \frac{du}{3}.
Also, 1 - x^6 = 1 - (x^3)^2 = 1 - u^2

Now substitute:
I = \int \frac{x^2}{1 - x^6} dx

I = \int \frac{1}{1 - u^2} \cdot \frac{du}{3}

I = \frac{1}{3} \int \frac{1}{1-u^2} du

Recall, \int \frac{1}{a^2 - x^2} \; dx = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C

So,
I = \frac{1}{2} \cdot \frac{1}{3} \int \log\left| \frac{1+u}{1-u} \right| du

I= \frac{1}{6} \log\left| \frac{1+u}{1-u} \right| + C

Substitute back u = x^3:

I = \frac{1}{6} \log\left| \frac{1 + x^3}{1 - x^3} \right| + C

Final Answer: \boxed{ \int \frac{x^2}{1 - x^6} dx = \frac{1}{6} \log\left| \frac{1 + x^3}{1 - x^3} \right| + C }

Q7. Evaluate \displaystyle \int \frac{x-1}{\sqrt{x^2 - 1}}; dx

Solution:
I = \int \frac{x-1}{\sqrt{x^2 - 1}} dx

Split the numerator:
I = \int \frac{x}{\sqrt{x^2 - 1}} dx - \int \frac{1}{\sqrt{x^2 - 1}} dx

I = I_1 - I_2

First term:
I_1 = \int \frac{x}{\sqrt{x^2 - 1}} dx

Let u = x^2 - 1 \implies du = 2x dx \implies x dx = \frac{du}{2}.

I_1 = \frac{1}{2} \int u^{-1/2} du

I_1 = u^{1/2} + C

I_1 = \sqrt{x^2 - 1} + C

Second term:
I_2 = \int \frac{1}{\sqrt{x^2 - 1}} dx

using: \int \frac{1}{\sqrt{x^2 - a^2}} \; dx = \log \left| x + \sqrt{x^2 - a^2} \right| + C

I_2 = \log\left| x + \sqrt{x^2 - 1} \right| + C

Substituting I_1 and I_2
I = \sqrt{x^2 - 1} - \log\left| x + \sqrt{x^2 - 1} \right| + C

Final Answer: \boxed{ \int \frac{x-1}{\sqrt{x^2 - 1}} dx = \sqrt{x^2 - 1} - \log\left| x + \sqrt{x^2 - 1} \right| + C }

Q8. Evaluate \displaystyle \int \frac{x^2}{\sqrt{x^6 + a^6}} dx

Solution:
I = \int \frac{x^2}{\sqrt{x^6 + a^6}} dx

Let u = x^3 \implies du = 3x^2 dx \implies x^2 dx = \frac{du}{3}.
So, x^6 + a^6 = (x^3)^2 + a^6 = u^2 + a^6

Now substitute:
I = \int \frac{1}{\sqrt{u^2 + a^6}} \cdot \frac{du}{3}

I = \frac{1}{3} \int \frac{1}{\sqrt{u^2 + a^6}} du

Use the formula: \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left| x + \sqrt{x^2 + a^2} \right| + C

I = \frac{1}{3} \log\left| u + \sqrt{u^2 + a^6} \right| + C

Substituting back u:
I = \frac{1}{3} \log\left| x^3 + \sqrt{x^6 + a^6} \right| + C

Final Answer: \boxed{ \int \frac{x^2}{\sqrt{x^6 + a^6}} dx = \frac{1}{3} \log\left| x^3 + \sqrt{x^6 + a^6} \right| + C }

Q9. Evaluate \displaystyle \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} ; dx

Solution:
I = \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx

Let u = \tan x \implies du = \sec^2 x dx.

Now substituting:
I = \int \frac{du}{\sqrt{u^2 + 4}}

Use the formula: \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left| x + \sqrt{x^2 + a^2} \right| + C

I = \log\left| u + \sqrt{u^2 + 4} \right| + C

Substituting back u:
I = \log\left| \tan x + \sqrt{\tan^2 x + 4} \right| + C

Final Answer: \boxed{ \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx = \log\left| \tan x + \sqrt{\tan^2 x + 4} \right| + C }

Q10. Evaluate \displaystyle \int \frac{1}{\sqrt{x^2 + 2x + 2}} ; dx

Solution:
I = \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx

Complete the square in the denominator:
x^2 + 2x + 2 = (x + 1)^2 + 1

So,
I = \int \frac{1}{\sqrt{(x + 1)^2 + 1}} dx

Use the formula: \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left| x + \sqrt{x^2 + a^2} \right| + C

Let, u = x + 1,
so, du = dx:

\int \frac{1}{\sqrt{u^2 + 1}} du = \log\left| u + \sqrt{u^2 + 1} \right| + C

Back substitute:
I = \log\left| x + 1 + \sqrt{x^2 + 2x + 2} \right| + C

\boxed{ \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx = \log\left| x + 1 + \sqrt{x^2 + 2x + 2} \right| + C }

Q11. Evaluate \displaystyle \int \frac{1}{9x^2 + 6x + 5} dx

Solution:
I = \int \frac{1}{9x^2 + 6x + 5} dx

First, complete the square in the denominator:
9x^2 + 6x + 5 \\ = 9\left(x^2 + \frac{2x}{3}\right) + 5 \\ = 9\left(x^2 + \frac{2x}{3} + \frac{1}{9}\right) + 5 - 1 \\ = 9\left(x + \frac{1}{3}\right)^2 + 4

I = \int \frac{1}{9\left(x + \frac{1}{3}\right)^2 + 4} dx

Let, u = x + \frac{1}{3}.
So, du = dx

So,
I = \int \frac{1}{9u^2 + 4} du

I = \int \frac{1}{9\left(u^2 + \frac{4}{9}\right)} du

I = \frac{1}{9} \int \frac{1}{u^2 + (\frac{2}{3})^2} du

Using: \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left( \frac{x}{a} \right) + C

Here, a = \frac{2}{3}:

I = \frac{1}{9} \cdot \frac{3}{2} \tan^{-1}\left( \frac{3u}{2} \right)

I = \frac{1}{6} \tan^{-1}\left( \frac{3u}{2} \right)

Substitute back u = x + \frac{1}{3}

I = \frac{1}{6} \tan^{-1}\left( \frac{3x + 1}{2} \right) + C

Final Answer: \boxed{ \int \frac{1}{9x^2 + 6x + 5} dx = \frac{1}{6} \tan^{-1}\left( \frac{3x + 1}{2} \right) + C }

Q12. Evaluate \displaystyle \int \frac{1}{\sqrt{7 - 6x - x^2}} ; dx

Solution:
Complete the square in the denominator:
7 - 6x - x^2 \\ = -(x^2 + 6x - 7) \\ = -(x^2 + 6x + 9 - 16) \\ = -\left( (x + 3)^2 - 16 \right) \\ = 16 - (x + 3)^2

I = \int \frac{1}{\sqrt{7 - 6x - x^2}} dx

I = \int \frac{1}{\sqrt{16 - (x + 3)^2}} dx

Here, substitute u = x + 3, so du = dx:

I = \int \frac{1}{\sqrt{16 - u^2}} du

Standard formula: \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left( \frac{x}{a} \right) + C

I = \sin^{-1} \left( \frac{u}{4} \right) + C

Substitute back u = x + 3:

I = \sin^{-1}\left( \frac{x + 3}{4} \right) + C

Final Answer: \boxed{ \int \frac{1}{\sqrt{7 - 6x - x^2}} dx = \sin^{-1}\left( \frac{x + 3}{4} \right) + C }

Q13. Evaluate \displaystyle \int \frac{1}{\sqrt{(x-1)(x-2)}} dx

Solution:
I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx

First, expand the denominator:
(x-1)(x-2) = x^2 - 3x + 2

Complete the square:
x^2 - 3x + 2 \\ = (x^2 - 3x + \frac{9}{4}) + 2 - \frac{9}{4} \\ = (x - \frac{3}{2})^2 - \frac{1}{4}

So, I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx

Using formula, \int \frac{1}{\sqrt{x^2 - a^2}} dx = \log\left| x + \sqrt{x^2 - a^2} \right| + C

Let u = x - \frac{3}{2}, a = \frac{1}{2}:

So, I = \int \frac{1}{\sqrt{(u)^2 - (\frac{1}{2})^2}} dx

Using formula:
I = \log\left| u + \sqrt{u^2 - (\frac{1}{2})^2} \right| + C

Substitute back u = x - \frac{3}{2}:

I = \log\left| x -\frac{3}{2} + \sqrt{(x-\frac{3}{2})^2 - (\frac{1}{2}^2)} \right|

I = \log\left| x - \frac{3}{2} + \sqrt{(x-1)(x-2)} \right| + C

\boxed{ \int \frac{1}{\sqrt{(x-1)(x-2)}} dx = \log\left| x - \frac{3}{2} + \sqrt{(x-1)(x-2)} \right| + C }

Q14. Evaluate \displaystyle \int \frac{1}{\sqrt{8 + 3x -x^2}} dx

Solution:
I = \int \frac{1}{\sqrt{8 + 3x -x^2}} dx

Arrange and complete the square in the denominator:
8 + 3x - x^2 = -x^2 + 3x + 8 = -(x^2 - 3x - 8)

x^2 - 3x - 8 \\ = x^2 - 3x + \frac{9}{4} - \frac{9}{4} - 8 \\ = (x - \frac{3}{2})^2 - \frac{41}{4}

8 + 3x - x^2 = \frac{41}{4} - (x - \frac{3}{2})^2

I =\int \frac{1}{\sqrt {\frac{41}{4} - (x - \frac{3}{2})^2 }} dx

Let u = x - \frac{3}{2},
so du = dx.

Now the integral becomes:

I = \int \frac{1}{\sqrt{\frac{41}{4} - u^2}} du

This matches the standard formula:
\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1} \left( \frac{x}{a} \right) + C

Here, a = \frac{\sqrt{41}}{2}, and x is replaced by u:

I = \sin^{-1} \left( \frac{2u}{\sqrt{41}} \right) + C

I = \sin^{-1} \left( \frac{2u}{\sqrt{41}} \right) + C

Now, recall u = x - \frac{3}{2},
so 2u = 2x - 3:

I = \sin^{-1} \left( \frac{2x - 3}{\sqrt{41}} \right) + C

Final Answer: \boxed{ \int \frac{1}{\sqrt{8 + 3x - x^2}} dx = \sin^{-1} \left( \frac{2x - 3}{\sqrt{41}} \right) + C }

Q15. Evaluate \int \frac{1}{\sqrt{(x-a)(x-b)}} dx

Solution:
Expand the denominator,
(x-a)(x-b) = x^2 - (a+b)x + ab

So the integral becomes,
\int \frac{1}{\sqrt{x^2 - (a+b)x + ab}} dx

Complete the square in the denominator,
x^2 - (a+b)x + ab \\ = \left[x^2 - (a+b)x + \frac{(a+b)^2}{4}\right] + ab - \frac{(a+b)^2}{4} \\ = \left[x - \frac{a+b}{2}\right]^2 + ab - \frac{(a+b)^2}{4} \\

Now,
ab - \frac{(a+b)^2}{4} \\ = ab - \frac{a^2 + 2ab + b^2}{4} \\ = ab - \frac{a^2}{4} - \frac{2ab}{4} - \frac{b^2}{4} \\ = ab - \frac{a^2}{4} - \frac{ab}{2} - \frac{b^2}{4} \\ = \frac{4ab - a^2 - 2ab - b^2}{4} \\ = \frac{2ab - a^2 - b^2}{4} \\

So the denominator is:
\left[x - \frac{a+b}{2}\right]^2 + \frac{2ab - a^2 - b^2}{4}

Recall, \int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C

So write:
\int \frac{1}{\sqrt{(x-a)(x-b)}} dx = \int \frac{1}{\sqrt{ \left( x - h \right)^2 + k^2 }} dx

where,
h = \frac{a+b}{2}, \quad k^2 = \frac{2ab - a^2 - b^2}{4}

Apply substitution,
Let u = x - \frac{a+b}{2}, so du = dx.

So,
\int \frac{1}{\sqrt{u^2 + k^2}} du = \log\left| u + \sqrt{u^2 + k^2} \right| + C

Substitute back:
I = \log\left| x - \frac{a+b}{2} + \sqrt{ (x-a)(x-b) } \right| + C

Final Answer: \boxed{ \int \frac{1}{\sqrt{(x-a)(x-b)}} dx = \log \left| x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right| + C }

Q16. Evaluate \displaystyle \int \frac{4x+1}{\sqrt{2x^2 + x - 3}} dx

Solution:
Let’s write the denominator in the form \sqrt{ax^2 + bx + c}.
Differentiate the inside:

\frac{d}{dx}(2x^2 + x - 3) = 4x + 1

So, set u = 2x^2 + x - 3.
Then, du = (4x + 1)dx.

So,
I = \int \frac{4x + 1}{\sqrt{2x^2 + x - 3}} dx

I = \int \frac{1}{\sqrt{u}} du

I = 2\sqrt{u} + C

Back substitute u = 2x^2 + x - 3:
Final Answer: \boxed{\int \frac{4x + 1}{\sqrt{2x^2 + x - 3}} dx = 2\sqrt{2x^2 + x - 3} + C }

Q17. Evaluate \displaystyle \int \frac{x+2}{\sqrt{x^2 - 1}} ; dx

Solution:
Split the numerator:
\int \frac{x}{\sqrt{x^2 - 1}} dx + \int \frac{2}{\sqrt{x^2 - 1}} dx

The first part:
Let u = x^2 - 1 \implies du = 2x dx \implies x dx = \frac{du}{2}

So,
\int \frac{x}{\sqrt{x^2 - 1}} dx = \frac{1}{2} \int u^{-1/2} du = u^{1/2} + C = \sqrt{x^2 - 1} + C

The second part:
2 \int \frac{1}{\sqrt{x^2 - 1}} dx = 2 \log \left| x + \sqrt{x^2 - 1} \right| + C

Combining,
Final Answer: \boxed{ \int \frac{x+2}{\sqrt{x^2 - 1}} dx = \sqrt{x^2 - 1} + 2\log \left| x + \sqrt{x^2 - 1} \right| + C }

Q18. Evaluate \displaystyle \int \frac{5x-2}{1 + 2x + 3x^2} dx

Solution:
Write the denominator in standard quadratic form
1 + 2x + 3x^2 = 3x^2 + 2x + 1

Express the numerator in terms of the derivative of the denominator
Differentiate the denominator:
\frac{d}{dx}(3x^2 + 2x + 1) = 6x + 2

Express 5x−2 as a linear combination of 6x+2 and a constant:
Let, 5x - 2 = a(6x + 2) + b
5x - 2 = 6ax + (2a + b)

Comparing coefficients:
For x: 5=6a  \implies a = \frac{5}{6}
​
For constant: −2=2a+b  ⟹  b=−2−2a
b = -2 -2 \frac{5}{6}
b = -2 - \frac{5}{3}
b = - \frac{11}{3}

So, 5x - 2 = \frac{5}{6}(6x + 2) - \frac{11}{3}

Substitute back into the integral
I = \int \frac{5x-2}{3x^2 + 2x + 1} dx

I= \int \frac{\frac{5}{6}(6x+2) - \frac{11}{3}}{3x^2 + 2x + 1} dx

Split the integral:
I = \frac{5}{6} \int \frac{6x + 2}{3x^2 + 2x + 1} dx - \frac{11}{3} \int \frac{1}{3x^2 + 2x + 1} dx

Solve the first integral

I_1 = \int \frac{6x + 2}{3x^2 + 2x + 1} dx

Let u = 3x^2 + 2x + 1 \implies \frac{du}{dx} = 6x + 2 \implies du = (6x + 2)dx

So, I_1 = \int \frac{6x+2}{3x^2 + 2x + 1} dx

I_1 = \int \frac{du}{u}

I_1 = \log |u| + C

I_1 = \log |3x^2 + 2x + 1| + C

Solve the second integral by completing the square
I_2 = \frac{11}{3} \int \frac{1}{3x^2 + 2x + 1} dx

First, complete the square in the denominator:
3x^2 + 2x + 1 = 3\left(x^2 + \frac{2}{3}x\right) + 1

Complete the square inside:
x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9}

So,
3\left(x + \frac{1}{3}\right)^2 - \frac{1}{3} + 1 = 3\left(x + \frac{1}{3}\right)^2 + \frac{2}{3}

3x^2 + 2x + 1 = 3\left(x + \frac{1}{3}\right)^2 + \frac{2}{3}

Therefore,
I_2 = \int \frac{1}{3\left(x + \frac{1}{3}\right)^2 + \frac{2}{3}} dx

Let y = x + \frac{1}{3}

Now,
I_2 = \int \frac{1}{3y^2 + \frac{2}{3}} dy

I_2 = \int \frac{1}{3\left(y^2 + \frac{2}{9}\right)} dy

I_2= \frac{1}{3} \int \frac{1}{y^2 + \left(\frac{\sqrt{2}}{3}\right)^2} dy

Recall the standard formula:
\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C

So,
I_2 = \frac{1}{3} \cdot \frac{3}{\sqrt{2}} \tan^{-1} \left( \frac{3y}{\sqrt{2}} \right)

I_2= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{3y}{\sqrt{2}} \right)

Substitute back, y= x + \frac{1}{3} :
I_2 = \frac {1}{\sqrt{2}} \tan^{-1} \left( \frac{3x + 1}{\sqrt{2}} \right) + C

Write the final answer

Combine both parts:
I = \frac{5}{6} \cdot I_1 - \frac{11}{3} \cdot I_2

\int \frac{5x-2}{3x^2 + 2x + 1} dx = \frac{5}{6} \log |3x^2 + 2x + 1| - \frac{11}{3\sqrt{2}} \tan^{-1} \left( \frac{3x + 1}{\sqrt{2}} \right) + C

Final Answer: \boxed{ \int \frac{5x-2}{1 + 2x + 3x^2} dx = \frac{5}{6} \log |3x^2 + 2x + 1| - \frac{11}{3\sqrt{2}} \tan^{-1} \left( \frac{3x + 1}{\sqrt{2}} \right) + C }

Q19. Evaluate \displaystyle \int \frac{6x + 7}{\sqrt{(x-5)(x-4)}} dx

Solution:
Expand and complete the square in the denominator:
(x-5)(x-4) = x^2 - 9x + 20

x^2 - 9x + 20 \\ = (x^2 - 9x + \frac{81}{4}) + 20 - \frac{81}{4} \\ = (x - \frac{9}{2})^2 - \frac{1}{4}

Now, let’s write numerator in terms of derivative of denominator:

\frac{d}{dx} (x^2 - 9x + 20) = 2x - 9

Write: 6x + 7 = A(2x - 9) + B
6x + 7 = 2Ax - 9A + B

Solving:
6x=2Ax  \implies  6x = 2A \implies A=3
7=−9A+B \implies  B = 7 + 27 \implies B = 34

So,
6x + 7 = 3(2x - 9) + 34

Split the integral:

I = \int \frac{6x + 7}{\sqrt{(x-5)(x-4)}} dx

I = 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx + \int \frac{34}{\sqrt{x^2 - 9x + 20}} dx

I = I_1 +I_2

First term:
I_1 = 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx

Let u=x^2−9x +20 \\ du=(2x - 9)dx

I_1 = 3 \int \frac{1}{\sqrt{u}} du

I_1= 3 \int u^{-1/2} du

I_1 = 6\sqrt{u} + C

I_1 = 6\sqrt{x^2 - 9x + 20}

Second term:
I_2 = \int \frac{34}{\sqrt{x^2 - 9x + 20}} dx

I_2 = 34 \int \frac{1}{\sqrt{(x - \frac{9}{2})^2 - (\frac{1}{2})^2}} dx

Using formula: \int \frac{1}{\sqrt{x^2 - a^2}} dx = \log \left| x + \sqrt{x^2 - a^2} \right| + C

So,
I_2 = 34 \int \frac{1}{\sqrt{(x - \frac{9}{2})^2 - (\frac{1}{2})^2}} dx

I_2 = 34 \log \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C

Combining the integrals:

I = I_1 + I_2
I = 6\sqrt{x^2 - 9x + 20} + 34 \log \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C

Final answer: \boxed{ \int \frac{6x + 7}{\sqrt{(x-5)(x-4)}} dx = 6\sqrt{x^2 - 9x + 20} + 34 \log \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C }

Q20. Evaluate \displaystyle \int \frac{x+2}{\sqrt{4x - x^2}} ; dx

Solution:
I = \int \frac{x+2}{\sqrt{4x - x^2}} dx

Rewrite denominator:
4x - x^2 = - (x^2 - 4x) = - \left( x^2 - 4x + 4 - 4 \right ) = - (x - 2)^2 + 4

So, \sqrt{4x - x^2} = \sqrt{4 - (x - 2)^2}

I = \int \frac{x+2}{\sqrt{4 - (x-2)^2}} dx

Let’s use the substitution u=x−2 \\ x=u+2, \\ dx=du

Then numerator: x+2=u+2+2=u+4

So the integral becomes:
I = \int \frac{u + 4}{\sqrt{4 - u^2}} du

I= \int \frac{u}{\sqrt{4 - u^2}} du + \int \frac{4}{\sqrt{4 - u^2}} du

I = I_1 +I_2

First term:
I_1 = \int \frac{u}{\sqrt{4 - u^2}} du

Let’s use substitution w = 4−u^2 \\ dw=−2u \cdot du \\ du = -\frac{1}{2} dw

I_1 = -\frac{1}{2} \int w^{-1/2} dw

I_1 = -w^{1/2} + C

I_1 = -\sqrt{4 - u^2}

Second term:

I_2 = 4 \int \frac{1}{\sqrt{4 - u^2}} du

Using the formula: \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1} \left( \frac{x}{a} \right) + C

I_2 = 4 \cdot \sin^{-1} \left( \frac{u}{2} \right ) + C

Combining the integrals,

I = I_1 +I_2

I = -\sqrt{4 - u^2} + 4 \cdot \sin^{-1} \left( \frac{u}{2} \right ) + C

Return to u=x−2

I = -\sqrt{4 - (x-2)^2} + 4 \cdot \sin^{-1} \left( \frac{x-2}{2} \right ) + C

Final Answer: \boxed{ \int \frac{x+2}{\sqrt{4x - x^2}} dx = -\sqrt{4x^2 - x^2} + 4 \sin^{-1} \left( \frac{x - 2}{2} \right) + C }

Q21. Evaluate \displaystyle \int \frac{x+2}{\sqrt{x^2 + 2x + 3}} dx

Solution:
I = \int \frac{x+2}{\sqrt{x^2 + 2x + 3}} dx

Complete the square in the denominator:
x^2 + 2x + 3 = (x + 1)^2 + 2

Let u = x+1 \\ x = u−1 \\ dx = du

Then numerator: x+2=u−1+2=u+1

So the integral becomes:
I = \int \frac{u + 1}{\sqrt{u^2 + 2}} du

I = \int \frac{u}{\sqrt{u^2 + 2}} du + \int \frac{1}{\sqrt{u^2 + 2}} du

I = I_1 + I_2

First term:
I_1 = \int \frac{u}{\sqrt{u^2 + 2}} du

let, w = u^2+2 \\ dw = 2udu \\ \frac{1}{2} \cdot dw = u \cdot du

I_1 = \frac{1}{2} \int w^{-1/2} dw

I_1 = w^{1/2} + C

I_1 = \sqrt{u^2 + 2}

Second term:
I_2 = \int \frac{1}{\sqrt{u^2 + 2}} du

I_2 = \log \left| u + \sqrt{u^2 + 2} \right| + C

Combining the integrals:
I = \sqrt{u^2 + 2} + \log \left| u + \sqrt{u^2 + 2} \right| + C

Return to: u=x+1
I = \int \frac{x+2}{\sqrt{x^2 + 2x + 3}} dx = \sqrt{x^2 + 2x + 3} + \log \left| x + 1 + \sqrt{x^2 + 2x + 3} \right| + C

Final Answer: \boxed{ \int \frac{x+2}{\sqrt{x^2 + 2x + 3}} dx = \sqrt{x^2 + 2x + 3} + \log \left| x + 1 + \sqrt{x^2 + 2x + 3} \right| + C }

Q22. Evaluate \displaystyle \int \frac{x+3}{x^2 - 2x - 5} dx

Solution:
I = \int \frac{x+3}{x^2 - 2x - 5} dx

Express numerator as A times the derivative of denominator plus B:
x + 3 = A(2x - 2) +B
x + 3 = 2Ax - 2A + B

Comparing coefficients,
\implies 1 = 2A \implies \frac{1}{2} = A

\implies 3 = -2A +B \implies 3 = -1 +B \implies 4 = B

We can write x+3=\frac{1}{2}(2x−2) + 4

So, I = \int \frac{x+3}{x^2 - 2x - 5} dx

I = \frac{1}{2} \int \frac{2x - 2}{x^2 - 2x - 5} dx + 4 \int \frac{1}{x^2 - 2x - 5} dx

I = I_1 + I_2

First term:
I_1 = \frac{1}{2} \int \frac{2x - 2}{x^2 - 2x - 5} dx

Let u=x^2−2x−5u \\ du=(2x−2)dx

I_1 = \frac{1}{2} \int \frac{du}{u}

I_1 = \frac{1}{2} \log|u| + C

I_1 = \frac{1}{2} \log|x^2 - 2x - 5| +C

Second term:
I_2 = 4 \int \frac{1}{x^2 - 2x - 5} dx

Complete the square in the denominator:
x^2 - 2x - 5 = (x - 1)^2 - 6

So,
I_2 = 4 \int \frac{1}{(x - 1)^2 - ( \sqrt{6} )^2} dx

Use the standard formula:
\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C

Here, let y=x−1
I_2 = 4 \int \frac{1}{y^2 - 6} dy

I_2 = 4 \cdot \frac{1}{2\sqrt{6}} \log \left| \frac{y - \sqrt{6}}{y + \sqrt{6}} \right|

I_2 = \frac{2}{\sqrt{6}} \log \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right|

Combining the integrals:
I = I_1 + I_2
I = \frac{1}{2} \log|x^2 - 2x - 5| + \frac{2}{\sqrt{6}} \log \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C

Final Answer: \boxed{ \int \frac{x+3}{x^2 - 2x - 5} dx = \frac{1}{2} \log|x^2 - 2x - 5| + \frac{2}{\sqrt{6}} \log \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C }

Q23. Evaluate \displaystyle \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} ; dx

Solution:
I = \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} ; dx

Let’s write the numerator in terms of the derivative of the denominator.
Differentiate denominator:
\frac{d}{dx}(x^2 + 4x + 10) = 2x + 4

Express 5x + 3 = A(2x + 4) + B:
5x +3 = 2Ax + 4A + B

Comparing coefficients
5x=2Ax  \implies 5 = 2A \implies  A=\frac{5}{2}

3=4A+B  \implies  B=3−4A \implies B=3−10 \implies B =-7

So, 5x + 3 = \frac{5}{2}(2x + 4) - 7

Split the integral:
I = \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx

I = \frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx - 7 \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx

I = I_1 +I_2

First term:
I_1 = \frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx

Let u = x^2 + 4x + 10,\ du = (2x + 4)dx:
I_1 = \frac{5}{2} \int \frac{du}{\sqrt{u}}

I_1 = \frac{5}{2} \cdot \frac{2}{1} \cdot u^{\frac{1}{2}} + C

I_1 = 5 \sqrt{u} + C

I_1 = 5\sqrt{x^2 + 4x + 10} + C

Second term:
I_2 = 7 \int \frac{1}{\sqrt{(x+2)^2 + 6}} dx

Let, x+2 = u \\ dx = du

I_2= 7 \int \frac{1}{\sqrt{u^2 + a^2}} du

Using Formula: \int \frac{1}{\sqrt{x^2 + a^2}} dx = \log\left| x + \sqrt{x^2 + a^2} \right| + C

I_2 = 7 \log\left| x + 2 + \sqrt{x^2 + 4x + 10} \right|

Final answer: \boxed{ \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx = 5 \sqrt{x^2 + 4x + 10} - 7 \log\left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C }

Q24. Evaluate \displaystyle \int \frac{dx}{x^2 + 2x + 2}

Options:
(A) x \tan^{-1} (x+1) + C
(B) \tan^{-1} (x+1) + C
(C) (x+1) \tan^{-1} x + C
(D) \tan^{-1} x +C

Solution:
I = \int \frac{dx}{x^2 + 2x + 2}

Complete the square in the denominator:
x^2 + 2x + 2 = (x + 1)^2 + 1

So,
I = \int \frac{dx}{(x + 1)^2 + 1}

Use the formula: \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C
Here, a=1

I = \tan^{-1}(x + 1) + C

Correct option: (B), \boxed{ \tan^{-1}(x + 1) + C }

Q25. Evaluate \displaystyle \int \frac{dx}{\sqrt{9x - 4x^2}}

Options:

(A) \frac{1}{9} \sin^{-1} \left( \frac{9x-8}{8} \right) + C

(B) \frac{1}{2} \sin^{-1} \left( \frac{8x-9}{9} \right) + C

(C) \sin^{-1} \left( \frac{9x-8}{8} \right) + C

(D) \sin^{-1} \left( \frac{9x-8}{9} \right) + C

Solution:
I = \int \frac{dx}{\sqrt{9x - 4x^2}}

First, rewrite denominator:
9x - 4x^2 = -4x^2 + 9x = -4\left( x^2 - \frac{9}{4}x \right)

Complete the square:
x^2 - \frac{9}{4}x = \left( x^2 - \frac{9}{4}x + \frac{81}{64} \right) - \frac{81}{64} = \left( x - \frac{9}{8} \right)^2 - \frac{81}{64}

So,
9x - 4x^2 = -4\left( (x - \frac{9}{8})^2 - \frac{81}{64} \right)

9x - 4x^2 = 4\left( \frac{81}{64} - (x - \frac{9}{8})^2 \right)

Therefore, \sqrt{9x - 4x^2} = 2 \sqrt{ \frac{81}{64} - (x - \frac{9}{8})^2 }

Let u= x - \frac{9}{8}, \implies du = dx

I = \int \frac{du}{2 \sqrt{ \frac{81}{64} - u^2 } }

I = \frac{1}{2} \int \frac{1}{ \sqrt{ \left( \frac{9}{8} \right)^2 - u^2 } } du

Using Formula: \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1} \left( \frac{x}{a} \right ) + C

So, here a=\frac{9}{8} and x is replaced by u:

I = \frac{1}{2} \sin^{-1} \left( \frac{u}{\frac{9}{8}} \right ) + C

I = \frac{1}{2} \sin^{-1} \left( \frac{8u}{9} \right ) + C

Recall, u = x - \frac{9}{8}

I = \frac{1}{2} \sin^{-1} \left( \frac{8x - 9}{9} \right ) + C

Correct Option (B): \boxed{\frac{1}{2} \sin^{-1} \left( \frac{8x - 9}{9} \right ) + C }