∫x5−xx4+11+x8dx=122tan−1⁡f(x)+C then f(x)=−−

I=∫x5−xx4+11+x8dx=12∫x4−1x4+11+x82xdx Let x2=t⇒2xdx=dtI=12∫t2−1t2+11+t4dtI=12∫t21−1t2t2t+1tt2+1t2dt Put t+1t=Z⇒1−1t2dt=dzt2+1t2=Z2−2I=12∫1zz2−2dzI=12∫zz2z2−2dz Let z2−2=y22zdz=2ydyI=12∫yy2+2y2dy=12∫1y2+(2)2dy=1212tan−1⁡y2=122tan−1⁡z2−22+C=122tan−1⁡t+1t2−22+C=122tan−1⁡t2+1t22+C=122tan−1⁡x4+1x42+C=122tan−1⁡1+x82x4+C

∫x5−xx4+11+x8dx=122tan−1⁡f(x)+C then f(x)=−−

I=∫x5−xx4+11+x8dx=12∫x4−1x4+11+x82xdx Let x2=t⇒2xdx=dtI=12∫t2−1t2+11+t4dtI=12∫t21−1t2t2t+1tt2+1t2dt Put t+1t=Z⇒1−1t2dt=dzt2+1t2=Z2−2I=12∫1zz2−2dzI=12∫zz2z2−2dz Let z2−2=y22zdz=2ydyI=12∫yy2+2y2dy=12∫1y2+(2)2dy=1212tan−1⁡y2=122tan−1⁡z2−22+C=122tan−1⁡t+1t2−22+C=122tan−1⁡t2+1t22+C=122tan−1⁡x4+1x42+C=122tan−1⁡1+x82x4+C

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$\int \frac{x^{5} - x}{(x^{4} + 1) \sqrt{1 + x^{8}}} d x = \frac{1}{2 \sqrt{2}} \tan^{- 1} f (x) + C then f (x) = - -$

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$\sqrt{\frac{1 + x^{8}}{2 x^{4}}}$

$\frac{1 + x^{4}}{2 x^{2}}$

$\frac{1 + x^{8}}{2 x^{4}}$

None

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x^{5} - x}{(x^{4} + 1) \sqrt{1 + x^{8}}} d x = \frac{1}{2} \int \frac{x^{4} - 1}{(x^{4} + 1) \sqrt{1 + x^{8}}} 2 x d x \\ Let x^{2} = t \Rightarrow 2 x d x = d t \\ I = \frac{1}{2} \int \frac{t^{2} - 1}{(t^{2} + 1) \sqrt{1 + t^{4}}} d t \end{array}$

$\begin{array}{l} I = \frac{1}{2} \int \frac{t^{2} (1 - \frac{1}{t^{2}})}{t^{2} (t + \frac{1}{t}) \sqrt{t^{2} + \frac{1}{t^{2}}}} d t \\ Put t + \frac{1}{t} = Z \Rightarrow (1 - \frac{1}{t^{2}}) d t = d z \\ t^{2} + \frac{1}{t^{2}} = Z^{2} - 2 \\ I = \frac{1}{2} \int \frac{1}{z \sqrt{z^{2} - 2}} d z \\ I = \frac{1}{2} \int \frac{z}{z^{2} \sqrt{z^{2} - 2}} d z \\ Let z^{2} - 2 = y^{2} \\ 2 z d z = 2 y d y \\ I = \frac{1}{2} \int \frac{y}{(y^{2} + 2) \sqrt{y^{2}}} d y \end{array}$

$\begin{array}{l} = \frac{1}{2} \int \frac{1}{y^{2} + (\sqrt{2})^{2}} d y \\ = \frac{1}{2} \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{y}{\sqrt{2}}) \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\sqrt{\frac{z^{2} - 2}{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\sqrt{\frac{{(t + \frac{1}{t})}^{2} - 2}{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\sqrt{\frac{(t^{2} + \frac{1}{t^{2}})}{2}}) + C \end{array}$

$\begin{aligned} = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\sqrt{\frac{x^{4} + \frac{1}{x^{4}}}{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\sqrt{\frac{1 + x^{8}}{2 x^{4}}}) + C \end{aligned}$