First slide

Methods of integration

Question

$\int \frac{x^{2} - 1}{(x^{2} + 1) \sqrt{1 + x^{4}}} d x is equal to$

Difficult

A

$2 \tan^{- 1} \frac{\sqrt{x^{2} + 1 / x^{2}}}{\sqrt{2}} + c$
B

$\frac{1}{2} \tan^{- 1} \frac{\sqrt{x^{2} + 1 / x^{2}}}{2} + c$
C

$\frac{1}{\sqrt{2}} t a n^{- 1} \frac{\sqrt{x^{2} + 1 / x^{2}}}{\sqrt{2}} + c$
D

$None of these$

Solution

$\begin{aligned} I = \int \frac{x^{2} - 1}{(x^{2} + 1) \sqrt{x^{4} + 1}} d x = \int \frac{x^{2} (1 - 1 / x^{2})}{x^{2} (x + 1 / x) \sqrt{x^{2} + 1 / x^{2}}} d x \\ = \int \frac{(1 - 1 / x^{2}) d x}{(x + 1 / x) \sqrt{(x + 1 / x)^{2} - 2}} \end{aligned}$
$Putting x + 1 / x = t, we have I = \int \frac{d t}{t \sqrt{t^{2} - 2}}$
$Again putting t^{2} - 2 = y^{2}, 2 t d t = 2 y d y,$
$I = \int \frac{y d y}{(y^{2} + 2) y} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{\sqrt{x^{2} + 1 / x^{2}}}{\sqrt{2}} + c$

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