Evaluate ∫$$x2+4x4+16dx$$

Correct option is 4122tan−1⁡x2−42x2+C

Questions

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

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122tan−1⁡x2−4x2+C

122tan−1⁡x−42x2+C

122tan−1⁡x2−422+C

122tan−1⁡x2−42x2+C

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detailed solution

Correct option is D

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 ∫x2+4x4+16dx

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