Anand Classes offers a free downloadable PDF of the NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.2 (Set-2), prepared with clear, step-by-step explanations focusing on integration by substitution and other integral methods. Click the print button to download study material and notes.
NCERT Question 13: Find the Integral
$$\int \frac{x^2}{(2+3x^3)^3} dx$$
Solution:
$$\int \frac{x^2}{(2+3x^3)^3} dx$$
Let $2+3x^3 = t \implies 9x^2 dx = dt \implies dx = \dfrac{dt}{9x^2}$
Substitute in the integral:
$$
\int \frac{x^2}{(2+3x^3)^3} dx = \int \frac{x^2}{t^3} \cdot \frac{dt}{9x^2} = \frac{1}{9} \int t^{-3} dt
$$
Integrate:
$$
\frac{1}{9} \cdot \frac{t^{-2}}{-2} + C
$$
Simplify:
$$
-\frac{1}{18} t^{-2} + C
$$
Substitute back $(t = 2+3x^3)$:
$$
\boxed{-\frac{1}{18} (2+3x^3)^{-2} + C}
$$
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NCERT Question 14: Find the Integral
$$\int \frac{1}{x(\log x)^m} dx \quad x>0$$
Solution:
$$\int \frac{1}{x(\log x)^m} dx $$
Let $\log x = t \implies \dfrac{1}{x} dx = dt$
Substitute in the integral:
$$
\int \frac{1}{x(\log x)^m} dx = \int \frac{dt}{t^m}
$$
Integrate:
$$
\int t^{-m} dt = \frac{t^{-m+1}}{-m+1} + C, \quad m \neq 1
$$
Substitute back $t = \log x$:
$$
\boxed{\frac{(\log x)^{-m+1}}{-m+1} + C, \quad m \neq 1}
$$
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NCERT Question 15: Find the Integral
$$\int \frac{x}{9-4x^2} dx$$
Solution:
$$\int \frac{x}{9-4x^2} dx$$
Let $9-4x^2 = t \implies -8x dx = dt \implies dx = \dfrac{dt}{-8x}$
Substitute in the integral:
$$
\int \frac{x}{9-4x^2} dx = \int \frac{x}{t} \cdot \frac{dt}{-8x} = -\frac{1}{8} \int \frac{1}{t} dt
$$
Integrate:
$$
-\frac{1}{8} \int \frac{1}{t} dt = -\frac{1}{8} \log|t| + C
$$
Substitute back $t = 9-4x^2$:
$$
\boxed{-\frac{1}{8} \log|9-4x^2| + C}
$$
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NCERT Question 16: Find the Integral
$$\int e^{2x+3} dx$$
Solution:
$$\int e^{2x+3} dx$$
Let $2x+3 = t \implies 2 dx = dt \implies dx = \dfrac{dt}{2}$
Substitute in the integral:
$$
\int e^{2x+3} dx = \int e^t \cdot \frac{dt}{2} = \frac{1}{2} \int e^t dt
$$
Integrate:
$$
\frac{1}{2} e^t + C
$$
Substitute back $t = 2x+3$:
$$
\boxed{\frac{1}{2} e^{2x+3} + C}
$$
Practice exponential substitution integrals with Anand Classes — detailed NCERT solutions for CBSE, JEE and competitive exams.
NCERT Question 17: Find the Integral
$$\int \frac{x}{e^{x^2}} dx$$
Solution:
$$\int \frac{x}{e^{x^2}} dx$$
Let $x^2 = t \implies 2x dx = dt \implies dx = \dfrac{dt}{2x}$
Substitute in the integral:
$$
\int \frac{x}{e^{x^2}} dx = \frac{1}{2} \int e^{-t} dt
$$
Integrate:
$$
\frac{1}{2} (-e^{-t}) + C = -\frac{1}{2} e^{-t} + C
$$
Substitute back $t = x^2$:
$$
\boxed{-\frac{1}{2} e^{-x^2} + C}
$$
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NCERT Question 18: Find the Integral
$$\int \frac{e^{\tan^{-1}x}}{1+x^2} dx$$
Solution:
$$\int \frac{e^{\tan^{-1}x}}{1+x^2} dx$$
Let $\tan^{-1}x = t \implies \dfrac{1}{1+x^2} dx = dt$
Substitute in the integral:
$$
\int \frac{e^{\tan^{-1}x}}{1+x^2} dx = \int e^t dt
$$
Integrate:
$$
e^t + C
$$
Substitute back $t = \tan^{-1}x$:
$$
\boxed{e^{\tan^{-1}x} + C}
$$
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NCERT Question 19: Find the Integral
$$\int \frac{e^{2x}-1}{e^{2x}+1} \; dx$$
Solution:
$$\int \frac{e^{2x}-1}{e^{2x}+1} \; dx$$
Dividing numerator and denominator by $e^x$:
$$
\frac{e^{2x}-1}{e^{2x}+1} = \frac{e^x – e^{-x}}{e^x + e^{-x}}
$$
Let $e^x + e^{-x} = t$, then, $(e^x – e^{-x}) dx = dt$
Substituting in the integral:
$$
\int \frac{e^x – e^{-x}}{e^x + e^{-x}} \; dx = \int \frac{dt}{t}
$$
Integrating:
$$
\log|t| + C
$$
Substitute back $t = e^x + e^{-x}$:
$$
\boxed{\log|e^x + e^{-x}| + C}
$$
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NCERT Question 20: Find the Integral
$$\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx$$
Solution:
$$\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx$$
Let $e^{2x}+e^{-2x}=t$
Then $dt=2(e^{2x}-e^{-2x})dx$
So $(e^{2x}-e^{-2x})dx=\frac{1}{2}dt$
Therefore
$$
\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx
= \int \frac{1}{t}\cdot\frac{1}{2}dt
= \frac{1}{2}\int \frac{1}{t}dt
$$
$$
= \frac{1}{2}\log|t|+C
$$
Substitute back $t=e^{2x}+e^{-2x}$
$$
\boxed{\frac{1}{2}\log|e^{2x}+e^{-2x}|+C}
$$
Master integration techniques with Anand Classes perfect for CBSE and JEE preparation
