Evaluate ∫x2+4x4+16dx

I=∫x2+4x4+16dx=∫1+4x2x2+16x2dx=∫1+4x2x2+4x2−8+8dx=∫1+4x2x−4x2+8dx Let x−4x=t . Then dx−4x=dt or 1+4x2dx=dt ∴ I=∫dtt2+(22)2=122tan−1⁡t22+C=122tan−1⁡x−4x22+C=122tan−1⁡x2−42x2+C

Evaluate ∫x2+4x4+16dx

I=∫x2+4x4+16dx=∫1+4x2x2+16x2dx=∫1+4x2x2+4x2−8+8dx=∫1+4x2x−4x2+8dx Let x−4x=t . Then dx−4x=dt or 1+4x2dx=dt ∴ I=∫dtt2+(22)2=122tan−1⁡t22+C=122tan−1⁡x−4x22+C=122tan−1⁡x2−42x2+C

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$Evaluate \int \frac{x^{2} + 4}{x^{4} + 16} d x$

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$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 4}{2 \sqrt{2}}) + C$

$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 4}{2 x \sqrt{2}}) + C$

$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x - 4}{2 x \sqrt{2}}) + C$

$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 4}{x \sqrt{2}}) + C$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x^{2} + 4}{x^{4} + 16} d x = \int \frac{1 + \frac{4}{x^{2}}}{x^{2} + \frac{16}{x^{2}}} d x \\ = \int \frac{1 + \frac{4}{x^{2}}}{x^{2} + {(\frac{4}{x})}^{2} - 8 + 8} d x \\ = \int \frac{1 + \frac{4}{x^{2}}}{{(x - \frac{4}{x})}^{2} + 8} d x \\ Let x - \frac{4}{x} = t . Then d (x - \frac{4}{x}) = d t or (1 + \frac{4}{x^{2}}) d x = d t \\ \begin{array}{l} ∴ I = \int \frac{d t}{t^{2} + (2 \sqrt{2})^{2}} = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{t}{2 \sqrt{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x - \frac{4}{x}}{2 \sqrt{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 4}{2 x \sqrt{2}}) + C \end{array} \end{array}$