If f(x)=∫((x^2+ ^2 x)/(1+x^2)) ^2 x d x and f(0)=0 , then f(1) equals

Correct answer: A. 1-(π)/(4) aligned & Let f(x)=∫((x^2+ ^2 x)/(1+x^2)) ^2 x d x & =∫ (x^2 ^2 x+( ^2 x)/( ^2 x))/(1+x^2)d x aligned aligned & =∫ (x^2 ^2 x+ ^2 x)/(1+x^2)d x & =∫ (x^2(1+ ^2 x)+ ^2 x)/(1+x^2)d x & =∫ (x^2+ ^2 x(1+x^2))/(1+x^2)d x & =∫ (x^2)/(1+x^2)d x+∫ ^2 x d x & =∫ (x^2+1-1)/(1+x^2) dx+∫( ^2 x-1 dx. & =∫1 d x-∫ (d x)/(1+x^2)+∫ ^2 x d x- x & =- ^ -1 x+ ^ -1 x+c & Given: f(0)=0 & f(0)=- ^ -1 0+ ^2+c & c=0 & f(x)=- ^ -1 x+ ^2 x aligned Now, f(1)=- ^ -1 (1)+ 1= 1- (π)/(4)

JEE Main 201219 MayOnline

Exam: JEE Main
Year: 2012
Date: 19 May
Mode: Online
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2012 · 19 May · Online · Mathematics · Indefinite Integration

If f(x)=∫((x^2+ ^2 x)/(1+x^2)) ^2 x d x and f(0)=0 , then f(1)...

f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x

and

f(0)=0

, then

f(1)

equals

Options

A $\tan 1-\frac{\pi}{4}$
B $\tan 1+1$
C $\frac{\pi}{4}$
D $1-\frac{\pi}{4}$

Correct answer

A. $\tan 1-\frac{\pi}{4}$

Step-by-step solution

\begin{aligned}& \text{ Let }f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\ & =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2}d x \end{aligned}

\begin{aligned}& =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2}d x \\ & =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2}d x \\ & =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2}d x \\ & =\int \frac{x^2}{1+x^2}d x+\int \tan ^2 x d x \\ & =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right.\\ & =\int1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\intd x \\ & =-\tan^{-1} x+\tan ^{-1} x+\mathrm{c}\\ & \text{ Given: }f(0)=0 \\ & \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\ & \Rightarrow c=0 \\ & \therefore f(x)=-\tan ^{-1} x+\tan ^2 x \end{aligned}

Now, f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-

\frac{\pi}{4}

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