Anand Classes provides a comprehensive, free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.4 (Set-1 Study Material), tailored for students who wish to strengthen their grasp of integrating standard functions. Each solution is explained step-by-step, aligning with the latest CBSE/NCERT curriculum and designed to build solid foundation for both board exams and engineering/medical entrance preparations. Click the print button to download study material and notes.
NCERT Question 1: Evaluate the integral
$$\int \frac{3x^2}{x^6 + 1} \; dx$$
Solution
$$\int \frac{3x^2}{x^6 + 1} \; dx$$
Let $x^3 = t$
So $3x^2 dx = dt$
$$\int \frac{3x^2}{x^6 + 1} dx = \int \frac{dt}{t^2 + 1}$$
$$= \tan^{-1} t + C$$
$$= \tan^{-1} (x^3) + C$$
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NCERT Question 2: Evaluate the integral
$$\int \frac{1}{\sqrt{1 + 4x^2}} dx$$
Solution
$$\int \frac{1}{\sqrt{1 + 4x^2}} dx$$
Let $2x = t$
So $2 dx = dt$
$$\int \frac{1}{\sqrt{1 + 4x^2}} dx = \frac{1}{2} \int \frac{1}{\sqrt{1 + t^2}} dt$$
$$= \frac{1}{2} \ln \left| t + \sqrt{t^2 + 1} \right| + C$$
$$= \frac{1}{2} \ln \left| 2x + \sqrt{4x^2 + 1} \right| + C$$
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NCERT Question 3: Evaluate the integral
$$\int \frac{1}{(\sqrt{2 – x})^2 + 1}\; dx$$
Solution
$$\int \frac{1}{(\sqrt{2 – x})^2 + 1}\; dx$$
Let $2 – x = t$
So $-dx = dt$
$$\int \frac{1}{\sqrt{(2 – x)^2} + 1}; dx = -\int \frac{1}{\sqrt{t^2 + 1}}; dt$$
$$= -\ln \left| t + \sqrt{t^2 + 1} \right| + C$$
$$= -\ln \left| 2 – x + \sqrt{(2 – x)^2 + 1} \right| + C$$
$$\boxed{=-\ln \left| 2 – x + \sqrt{(2 – x)^2 + 1} \right| + C}$$
$$\boxed{=\ln \left|\frac{1}{ 2 – x + \sqrt{(2 – x)^2 + 1}}\right| + C}$$
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NCERT Question 4: Evaluate the integral
$$\int \frac{1}{\sqrt{9 – 25x^2}}\; dx$$
Solution
$$\int \frac{1}{\sqrt{9 – 25x^2}}\; dx$$
Let $5x = t$
So $5 dx = dt$
$$\int \frac{1}{\sqrt{9 – 25x^2}}\; dx = \frac{1}{5} \int \frac{1}{\sqrt{9 – t^2}}\; dt$$
$$= \frac{1}{5} \sin^{-1} \left( \frac{t}{3} \right) + C$$
$$= \frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C$$
$$\boxed{\frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C}$$
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NCERT Question 5: Evaluate the integral
$$\int \frac{3x}{1 + 2x^4}\ dx$$
Solution
$$\int \frac{3x}{1 + 2x^4}\ dx$$
Let $u = x^2$
So $du = 2x\ dx$ and $x\ dx = \dfrac{1}{2} du$
$$\int \frac{3x}{1 + 2x^4}\ dx = 3\int \frac{x\ dx}{1 + 2(x^2)^2}$$
$$= 3\int \frac{x\ dx}{1 + 2u^2}$$
$$= 3\int \frac{\dfrac{1}{2} du}{1 + 2u^2}$$
$$= \frac{3}{2}\int \frac{du}{1 + 2u^2}$$
Write $1 + 2u^2 = 1 + (\sqrt{2}\ u)^2$
$$\int \frac{du}{1 + 2u^2} = \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\ u) + C$$
Therefore
$$\frac{3}{2}\int \frac{du}{1 + 2u^2} = \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\ u) + C$$
Substitute $u = x^2$
$$= \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\ x^2) + C$$
$$\boxed{\frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\ x^2) + C}$$
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NCERT Question 6: Evaluate the integral
$$\int \frac{x^2}{1 – x^6}\ dx$$
Solution
$$\int \frac{x^2}{1 – x^6}\ dx$$
Let $u = x^3$
So $du = 3x^2\ dx$ and $x^2\ dx = \dfrac{1}{3}\ du$
$$\int \frac{x^2}{1 – x^6}\ dx = \frac{1}{3}\int \frac{du}{1 – u^2}$$
$$\frac{1}{3}\int \frac{du}{1 – u^2} = \frac{1}{3}\cdot\frac{1}{2}\ln\left|\frac{1 + u}{1 – u}\right| + C$$
$$= \frac{1}{6}\ln\left|\frac{1 + x^3}{1 – x^3}\right| + C$$
$$\boxed{\frac{1}{6}\ln\left|\frac{1 + x^3}{1 – x^3}\right| + C}$$
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NCERT Question 7: Evaluate the integral
$$\int \frac{x – 1}{\sqrt{x^2 – 1}}\ dx$$
Solution
$$\int \frac{x – 1}{\sqrt{x^2 – 1}}\ dx$$
$$= \int \frac{x}{\sqrt{x^2 – 1}}\ dx \;-\; \int \frac{1}{\sqrt{x^2 – 1}}\ dx$$
Let $u = x^2 – 1$
So $du = 2x\ dx$ and $x\ dx = \dfrac{1}{2}\ du$
$$\int \frac{x}{\sqrt{x^2 – 1}}\ dx = \frac{1}{2}\int \frac{du}{\sqrt{u}} = \frac{1}{2}\cdot 2\sqrt{u} = \sqrt{u} = \sqrt{x^2 – 1}$$
$$\int \frac{1}{\sqrt{x^2 – 1}}\ dx = \ln\left|x + \sqrt{x^2 – 1}\right| + C$$
Therefore
$$\int \frac{x – 1}{\sqrt{x^2 – 1}}\ dx = \sqrt{x^2 – 1} – \ln\left|x + \sqrt{x^2 – 1}\right| + C$$
$$\boxed{\sqrt{x^2 – 1} – \ln\left|x + \sqrt{x^2 – 1}\right| + C}$$
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NCERT Question 8: Evaluate the integral
$$\int \frac{x^2}{\sqrt{x^6 + a^6}}\ dx$$
Solution
$$\int \frac{x^2}{\sqrt{x^6 + a^6}}\ dx$$
Let $u = x^3$
So $du = 3x^2\ dx$ and $x^2\ dx = \dfrac{1}{3}\ du$
$$\int \frac{x^2}{\sqrt{x^6 + a^6}}\ dx = \frac{1}{3}\int \frac{du}{\sqrt{u^2 + (a^3)^2}}$$
$$= \frac{1}{3}\ln\left|u + \sqrt{u^2 + (a^3)^2}\right| + C$$
$$= \frac{1}{3}\ln\left|x^3 + \sqrt{x^6 + a^6}\right| + C$$
$$\boxed{\frac{1}{3}\ln\left|x^3 + \sqrt{x^6 + a^6}\right| + C}$$
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NCERT Question 9: Evaluate the integral
$$\int \frac{\sec^{2}x}{\sqrt{\tan^{2}x + 4}}\ dx$$
Solution
$$\int \frac{\sec^{2}x}{\sqrt{\tan^{2}x + 4}}\ dx$$
Let
$$u = \tan x$$
So
$$du = \sec^{2}x\ dx$$
Therefore
$$\int \frac{\sec^{2}x}{\sqrt{\tan^{2}x + 4}}\ dx = \int \frac{du}{\sqrt{u^{2} + 4}}$$
Write
$$\sqrt{u^{2} + 4} = \sqrt{u^{2} + 2^{2}}$$
Standard form:
$$\int \frac{du}{\sqrt{u^{2} + a^{2}}} = \ln\left|u + \sqrt{u^{2} + a^{2}}\right| + C$$
So
$$\int \frac{du}{\sqrt{u^{2} + 4}} = \ln\left|u + \sqrt{u^{2} + 2^2}\right| + C$$
Substitute back $u = \tan x$
$$= \ln\left|\tan x + \sqrt{\tan^{2}x + 4}\right| + C$$
$$\boxed{\ln\left|\tan x + \sqrt{\tan^{2}x + 4}\right| + C}$$
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NCERT Question 10: Evaluate the integral
$$\int \frac{1}{\sqrt{x^2 + 2x + 2}}\ dx$$
Solution
$$\int \frac{1}{\sqrt{x^2 + 2x + 2}}\ dx$$
Complete the square:
$$x^2 + 2x + 2 = (x + 1)^2 + 1$$
So the integral becomes:
$$\int \frac{1}{\sqrt{(x + 1)^2 + 1}}\ dx$$
Let
$$u = x + 1$$
So
$$du = dx$$
Then
$$\int \frac{1}{\sqrt{u^2 + 1}}\ du = \ln\left|u + \sqrt{u^2 + 1}\right| + C$$
Substitute back $(u = x + 1)$:
$$= \ln\left|x + 1 + \sqrt{(x + 1)^2 + 1}\right| + C$$
$$\boxed{\ln\left|x + 1 + \sqrt{(x + 1)^2 + 1}\right| + C}$$
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NCERT Question 11: Evaluate the integral
$$\int \frac{1}{9x^2 + 6x + 5}\ dx$$
Solution
$$\int \frac{1}{9x^2 + 6x + 5}\ dx$$
Complete the square
$$9x^2 + 6x + 5 = 9\left(x^2 + \frac{2}{3}x\right) + 5$$
$$= 9\left[\left(x + \frac{1}{3}\right)^2 – \frac{1}{9}\right] + 5$$
$$= 9\left(x + \frac{1}{3}\right)^2 – 1 + 5$$
$$= 9\left(x + \frac{1}{3}\right)^2 + 4$$
Let $$u = x + \frac{1}{3}$$ so $$du = dx$$
$$\int \frac{1}{9x^2 + 6x + 5}\ dx = \int \frac{1}{9u^2 + 4}\ du$$
$$= \frac{1}{9}\int \frac{1}{u^2 + \left(\dfrac{2}{3}\right)^2}\ du$$
Using $$\int \frac{1}{u^2 + a^2}\ du = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$$ with $$a = \dfrac{2}{3}$$
$$\frac{1}{9}\cdot\frac{1}{\dfrac{2}{3}}\tan^{-1}\left(\frac{u}{\dfrac{2}{3}}\right) + C = \frac{1}{9}\cdot\frac{3}{2}\tan^{-1}\left(\frac{3u}{2}\right) + C$$
$$= \frac{1}{6}\tan^{-1}\left(\frac{3u}{2}\right) + C$$
Substitute back $$u = x + \frac{1}{3}$$
$$= \frac{1}{6}\tan^{-1}\left(\frac{3x + 1}{2}\right) + C$$
$$\boxed{\frac{1}{6}\tan^{-1}\left(\frac{3x + 1}{2}\right) + C}$$
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NCERT Question 12: Evaluate the integral
$$\int \frac{1}{\sqrt{7 – 6x – x^{2}}}\ dx$$
Solution
$$\int \frac{1}{\sqrt{7 – 6x – x^{2}}}\ dx$$
Rewrite the quadratic in proper form:
$$7 – 6x – x^{2} = -(x^{2} + 6x – 7)$$
Complete the square:
$$x^{2} + 6x – 7 = (x + 3)^{2} – 9 – 7 = (x + 3)^{2} – 16$$
Thus
$$7 – 6x – x^{2} = -[(x + 3)^{2} – 16] = 16 – (x + 3)^{2}$$
So the integral becomes:
$$\int \frac{1}{\sqrt{16 – (x + 3)^{2}}}\ dx$$
Let
$$u = x + 3$$
So
$$du = dx$$
Then
$$\int \frac{1}{\sqrt{16 – u^{2}}}\ du = \sin^{-1}\left(\frac{u}{4}\right) + C$$
Substitute back $(u = x + 3)$:
$$= \sin^{-1}\left(\frac{x + 3}{4}\right) + C$$
$$\boxed{\sin^{-1}\left(\frac{x + 3}{4}\right) + C}$$
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