Indefinite Integration JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Indefinite Integration, free to read — no sign-in needed.

1. Q1JEE Advanced Adv 2012 (Paper 1)
   The integral $\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{\frac{9}{2}}} d x$ equals (for some arbitrary constant $\mathrm{K}$ )
   1. A.$-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
   2. B.$\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
   3. C.$-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
   4. D.$\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
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   Answer: (C)

   $\begin{array}{l} I=\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x \\ \text { Let } \sec x+\tan x=t \Rightarrow \sec x-\tan x=\frac{1}{t} \\ \Rightarrow \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right) \text { and } \sec x(\sec x+\tan x) d x=d t \\ \Rightarrow \sec x d x=\frac{d t}{t} \\ \therefore \quad I=\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right) d t}{t^{9 / 2} \cdot t}=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t \\ =\frac{-1}{7} t^{-7 / 2}-\frac{1}{11} t^{-1 / 2}+K \\ =-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+K=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^{2}}{7}\right)+K \\ =\frac{-1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K \end{array}$
2. Q2JEE Advanced Adv 2008 (Paper 2)
   Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$. Then, for an arbitrary constant $C$, the value of $J-I$ equals
   1. A.$\frac{1}{2} \log \left|\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right|+C$
   2. B.$\frac{1}{2} \log \left|\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right|+C$
   3. C.$\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$
   4. D.$\frac{1}{2} \log \left|\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right|+C$
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   Answer: (C)

   Since, $J=\int \frac{e^{3 x}}{1+e^{2 x}+e^{4 x}} d x$ $$\begin{aligned} \therefore J-I & =\int \frac{\left(e^{3 x}-e^x\right)}{1+e^{2 x}+e^{4 x}} d x=\int \frac{\left(u^2-1\right)}{1+u^2+u^4} d u \\ & =\int \frac{\left(1-\frac{1}{u^2}\right)}{1+\frac{1}{u^2}+u^2} d u=\int \frac{\left(1-\frac{1}{u^2}\right)}{\left(u+\frac{1}{u}\right)^2-1} d u \\ & =\int \frac{d t}{t^2-1} \\ & =\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C \\ & =\frac{1}{2} \log \left|\frac{u^2-u+1}{u^2+u+1}\right|+C=\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C \end{aligned}$$
3. Q3JEE Advanced Adv 2007 (Paper 2)
   Let $f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}}$ for $n \geq 2$ and $g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text { occurs } n \text { times }}(x)$. Then, $\int x^{n-2} g(x) d x$ equals
   1. A.$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$
   2. B.$\frac{1}{n-1}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$
   3. C.$\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$
   4. D.$\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$
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   Answer: (A)

   Here, $\begin{array}{rlrl} & \text { and } & f f(x) & =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ & \therefore \quad f f f(x) & =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \text { Let } & g(x) & =\underbrace{(f \circ f o \ldots o f)}_{n \text { times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I & =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ & & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array}$ $\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$.
4. Q4JEE Advanced Adv 2007 (Paper 1)
   Let $F(x)$ be an indefinite integral of $\sin ^2 x$. Statement I The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. Statement II $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.
   1. A.Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
   2. B.Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
   3. C.Statement I is true, Statement II is false
   4. D.Statement I is false, Statement II is true
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   Answer: (D)

   $$\begin{aligned} & F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2} d x \\ & \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c \end{aligned}$$ Since, $F(x+\pi) \neq F(x)$. Hence, Statement I is false. But Statement II is true as $\sin ^2 x$ is periodic with period $\pi$.
5. Q5JEE Advanced Adv 2006
   The value of $\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$ is
   1. A.$2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$
   2. B.$2 \sqrt{2+\frac{2}{x^2}+\frac{1}{x^4}}+C$
   3. C.$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$
   4. D.None of these
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   Answer: (C)

   Let $\quad I=\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$, On dividing numerater and denominator by $x^5$, we get $$=\int \frac{\left(\frac{1}{x^3}-\frac{1}{x^5}\right) d x}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$$ put $2-\frac{2}{x^2}+\frac{1}{x^4}=t \Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right) d x=d t$ $$\therefore \quad I=\frac{1}{4} \int \frac{d t}{\sqrt{t}}=\frac{1}{4} \cdot \frac{t^{1 / 2}}{1 / 2}=\frac{1}{2} \sqrt{t}+c=\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$$