Anand Classes offers a comprehensive free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.2 (Set-1), created by expert educators at Anand Classes to guide you through the method of substitution and other core integration techniques aligned with the latest CBSE/NCERT syllabus. Click the print button to download study material and notes.
NCERT Question 1: Find Integral
$$\int \frac{2x}{1+x^2}dx$$
Solution:
$$\int \frac{2x}{1+x^2}dx$$
Let $1 + x^2 = t$
Then, $2x,dx = dt$
Substitute in the integral:
$$
\int \frac{2x}{1+x^2}dx = \int \frac{1}{t}dt
$$
$$
= \log|t| + C
$$
Substituting back $t = 1 + x^2$:
$$
\boxed{\log|1 + x^2| + C}
$$
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NCERT Question 2: Find the Integral
$$\int \frac{(\log|x|)^2}{x}dx$$
Solution:
$$\int \frac{(\log|x|)^2}{x}dx$$
Let $\log|x| = t$
Then, $\dfrac{1}{x}dx = dt$
Substitute in the integral:
$$
\int \frac{(\log|x|)^2}{x}dx = \int t^2 dt
$$
$$
= \frac{t^3}{3} + C
$$
Substituting back $t = \log|x|$:
$$
\boxed{\frac{(\log|x|)^3}{3} + C}
$$
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NCERT Question 3: Find the Integral
$$\int \frac{1}{x + x\log x}dx$$
Solution:
$$\int \frac{1}{x + x\log x}dx$$
Simplify the integrand:
$$
\frac{1}{x + x\log x} = \frac{1}{x(1 + \log x)}
$$
Let $1 + \log x = t$
Then, $\dfrac{1}{x}dx = dt$
Substitute in the integral:
$$
\int \frac{1}{x(1 + \log x)}dx = \int \frac{1}{t} dt
$$
$$
= \log|t| + C
$$
Substituting back $t = 1 + \log x$:
$$
\boxed{\log|1 + \log x| + C}
$$
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NCERT Question 4: Find the Integral
$$\int \sin x \;\sin(\cos x)dx$$
Solution:
$$\int \sin x \;\sin(\cos x)dx$$
Let $\cos x = t$
Then, $- \sin x dx = dt$
Substitute in the integral:
$$
\int \sin x \;\sin(\cos x)dx = -\int \sin t dt
$$
$$
= -(-\cos t) + C
$$
$$
= \cos t + C
$$
Substituting back $t = \cos x$:
$$
\boxed{\cos(\cos x) + C}
$$
Master trigonometric substitution integrals with Anand Classes — complete NCERT solutions and step-by-step guidance for JEE, CBSE, and competitive exams.
NCERT Question 5: Find the Integral
$$\int \sin(ax+b)\cos(ax+b)dx$$
Solution:
$$\int \sin(ax+b)\cos(ax+b)dx$$
Use the identity:
$$
\sin(ax+b)\cos(ax+b) = \frac{1}{2}\sin 2(ax+b)
$$
Therefore,
$$
\int \sin(ax+b)\cos(ax+b)dx = \int \frac{1}{2} \sin 2(ax+b)dx
$$
Let $2(ax+b) = t$
Then, $2adx = dt \implies dx = \dfrac{dt}{2a}$
Substitute:
$$
\int \frac{1}{2} \sin 2(ax+b)dx = \frac{1}{2} \int \sin t \frac{dt}{2a} = \frac{1}{4a} \int \sin tdt
$$
$$
= \frac{1}{4a}(-\cos t) + C
$$
Substitute back $t = 2(ax+b)$:
$$
\boxed{-\frac{1}{4a}\cos 2(ax+b) + C}
$$
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NCERT Question 6: Find the Integral
$$\int \sqrt{ax+b}\;dx$$
Solution:
$$\int \sqrt{ax+b}\;dx$$
Let $ax+b = t$
Then, $a dx = dt \implies dx = \dfrac{dt}{a}$
Substitute in the integral:
$$
\int \sqrt{ax+b}\;dx = \frac{1}{a} \int t^{1/2} dt
$$
Integrate:
$$
\frac{1}{a} \cdot \frac{t^{3/2}}{3/2} + C
$$
Simplify:
$$
\frac{2}{3a} t^{3/2} + C
$$
Substitute back $t = ax+b$:
$$
\boxed{\frac{2}{3a} (ax+b)^{3/2} + C}
$$
Master substitution method integrals with Anand Classes — step-by-step NCERT solutions for CBSE, JEE, and competitive exams.
NCERT Question 7: Find the Integral
$$\int x\sqrt{x+2}\;dx$$
Solution:
$$\int x\sqrt{x+2}\;dx$$
Let $x+2 = t$
Then $dx = dt$
Substitute in the integral:
$$
\int x\sqrt{x+2}\;dx = \int (t-2)\sqrt{t}\;dt
$$
$$
= \int (t^{3/2} – 2t^{1/2})\;dt
$$
Integrate each term:
$$
\int t^{3/2} dt – 2\int t^{1/2} dt = \frac{t^{5/2}}{5/2} – 2\cdot \frac{t^{3/2}}{3/2} + C
$$
Simplify the fractions:
$$
\frac{2}{5} t^{5/2} – \frac{4}{3} t^{3/2} + C
$$
Substitute back $t = x+2$:
$$
\boxed{\frac{2}{5} (x+2)^{5/2} – \frac{4}{3} (x+2)^{3/2} + C}
$$
Master substitution and power rule integrals with Anand Classes — complete NCERT solutions for CBSE JEE and competitive exams.
NCERT Question 8: Find the Integral
$$\int x\sqrt{1+2x^2}\;dx$$
Solution:
$$\int x\sqrt{1+2x^2}\;dx$$
Let $1+2x^2 = t$
Then $4x\;dx = dt \implies dx = \dfrac{dt}{4x}$
Substitute in the integral:
$$
\int x\sqrt{1+2x^2}\;dx = \int \frac{\sqrt{t}}{4} dt
$$
Integrate:
$$
\frac{1}{4} \int t^{1/2} dt = \frac{1}{4} \cdot \frac{t^{3/2}}{3/2} + C
$$
Simplify:
$$
\frac{1}{6} t^{3/2} + C
$$
Substitute back $t = 1+2x^2$:
$$
\boxed{\frac{1}{6} (1+2x^2)^{3/2} + C}
$$
Master substitution and power rule integration with Anand Classes — complete NCERT solutions for CBSE JEE and competitive exams.
NCERT Question 9: Find the Integral
$$\int \frac{4x+2}{x^2+x+1}dx$$
Solution:
$$\int \frac{4x+2}{x^2+x+1}dx$$
Let $x^2 + x + 1 = t$
Then $ (2x+1)dx = dt$
Notice $4x+2 = 2(2x+1)$, so:
$$
\int \frac{4x+2}{x^2+x+1}dx = \int \frac{2(2x+1)}{t} dx = 2 \int \frac{dt}{t}
$$
$$
= 2 \int \frac{1}{t} dt = 2\log|t| + C
$$
Substitute back $t = x^2 + x + 1$:
$$
\boxed{2\log|x^2 + x + 1| + C}
$$
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NCERT Question 10: Find the Integral
$$\int \frac{1}{x-\sqrt{x}}dx$$
Solution:
$$\int \frac{1}{x-\sqrt{x}}dx$$
Write the above equation as :
$$\int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx$$
Let $\sqrt{x} = t \implies x = t^2 \implies dx = 2t\;dt$
Substitute in the integral:
$$
\int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx = \int \frac{1}{t(t-1)} \cdot 2t\;dt
$$
Simplify:
$$
\int \frac{2t}{t(t-1)} dt = \int \frac{2}{t-1} dt
$$
Integrate:
$$
\int \frac{2}{t-1} dt = 2\log|t-1| + C
$$
Substitute back $t = \sqrt{x}$:
$$
\boxed{2\log|\sqrt{x}-1| + C}
$$
If you want, I can also rewrite it in the exact NCERT style with substitution steps like the previous questions for full clarity. Do you want me to do that?
NCERT Question 11: Find the Integral
$$\int \frac{x}{\sqrt{x+4}} dx \quad x>0$$
Solution:
$$\int \frac{x}{\sqrt{x+4}} dx \quad x>0$$
Let $x+4 = t \implies dx = dt$
Then $x = t – 4$, so the integral becomes:
$$
\int \frac{x}{\sqrt{x+4}} dx = \int \frac{t-4}{\sqrt{t}} dt
$$
Split the terms:
$$
\int \frac{t}{\sqrt{t}} dt – \int \frac{4}{\sqrt{t}} dt = \int t^{1/2} dt – 4 \int t^{-1/2} dt
$$
Integrate each term:
$$
\frac{2}{3} t^{3/2} – 4 \cdot 2 t^{1/2} + C
$$
Simplify:
$$
\frac{2}{3} t^{3/2} – 8 t^{1/2} + C
$$
Substitute back $t = x+4$:
$$
\boxed{\frac{2}{3} (x+4)^{3/2} – 8 (x+4)^{1/2} + C}
$$
$$
\boxed{\frac{2}{3} \sqrt{x+4} (x-8) + C}
$$
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NCERT Question 12: Find the Integral
$$\int (x^3-1)^{1/3} x^5 dx$$
Solution:
$$\int (x^3-1)^{1/3} x^5 dx$$
Let $x^3-1 = t \implies 3x^2 dx = dt \implies dx = \dfrac{dt}{3x^2}$
Rewrite $x^5 dx$ as $x^3 \cdot x^2 dx$, then $x^2 dx = \dfrac{dt}{3}$:
$$
\int (x^3-1)^{1/3} x^5 dx = \int (t)^{1/3} \cdot x^3 \cdot x^2 dx = \frac{1}{3} \int t^{1/3} (t+1) dt
$$
Expand the integrand:
$$
\frac{1}{3} \int (t^{4/3} + t^{1/3}) dt
$$
Integrate each term:
$$
\frac{1}{3} \left( \frac{3}{7} t^{7/3} + \frac{3}{4} t^{4/3} \right) + C
$$
Simplify:
$$
\frac{1}{7} t^{7/3} + \frac{1}{4} t^{4/3} + C
$$
Substitute back $t = x^3-1$:
$$
\boxed{\frac{1}{7} (x^3-1)^{7/3} + \frac{1}{4} (x^3-1)^{4/3} + C}
$$
Master substitution and power rule integration with Anand Classes — detailed NCERT solutions for CBSE, JEE and competitive exams.
