∫x2−1x2+11+x4dx is equal to

I=∫x2−1x2+1x4+1dx=∫x21−1/x2x2(x+1/x)x2+1/x2dx=∫1−1/x2dx(x+1/x)(x+1/x)2−2 Putting x+1/x=t, we have I=∫dttt2−2 Again putting t2−2=y2,2tdt=2ydy, I=∫ydyy2+2y=12tan−1⁡y2=12tan−1⁡x2+1/x22+c

∫x2−1x2+11+x4dx is equal to

I=∫x2−1x2+1x4+1dx=∫x21−1/x2x2(x+1/x)x2+1/x2dx=∫1−1/x2dx(x+1/x)(x+1/x)2−2 Putting x+1/x=t, we have I=∫dttt2−2 Again putting t2−2=y2,2tdt=2ydy, I=∫ydyy2+2y=12tan−1⁡y2=12tan−1⁡x2+1/x22+c

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

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

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

$None of these$

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

answer is C.

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Detailed Solution

$\begin{aligned} I = \int \frac{x^{2} - 1}{(x^{2} + 1) \sqrt{x^{4} + 1}} d x = \int \frac{x^{2} (1 - 1 / x^{2})}{x^{2} (x + 1 / x) \sqrt{x^{2} + 1 / x^{2}}} d x \\ = \int \frac{(1 - 1 / x^{2}) d x}{(x + 1 / x) \sqrt{(x + 1 / x)^{2} - 2}} \end{aligned}$

$Putting x + 1 / x = t, we have I = \int \frac{d t}{t \sqrt{t^{2} - 2}}$

$Again putting t^{2} - 2 = y^{2}, 2 t d t = 2 y d y,$

$I = \int \frac{y d y}{(y^{2} + 2) y} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{\sqrt{x^{2} + 1 / x^{2}}}{\sqrt{2}} + c$