Let and f (0) = 0, then the value of f

JEE MainMathematicsModerateIndefinite integration

Let
$f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)}$
and f (0) = 0, then the value of f

Show answer and solution

f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)}

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Let

x = tan\theta \Rightarrow dx = \sec^{2}\theta d\theta

\Rightarrow f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)} = \int\frac{\tan^{2}\theta\sec^{2}\theta d\theta}{\sec^{2}\theta(1 + sec\theta)}

= \int\frac{\tan^{2}\theta d\theta}{1 + sec\theta}\theta = \int\frac{\sin^{2}\theta d\theta}{cos\theta(1 + cos\theta)}

= \int\frac{1 - \cos^{2}\theta d\theta}{cos\theta(1 + cos\theta)} = \int\frac{(1 - cos\theta)d\theta}{cos\theta} = \int sec\theta d\theta - \int d\theta

= log(x + \sqrt{1 + x^{2}}) - \tan^{- 1}x + c

Given f (0) = 0

\Rightarrow

0 = log⁡1 - 0 + c

\Rightarrow

c = 0

\Rightarrow

f(1) = log\left( 1 + \sqrt{1 + 1^{2}} \right) - \tan^{- 1}(1) = log(1 + \sqrt{2}) - \frac{\pi}{4}

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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