∫x4+1x6+1dx=Atan-1x+1Btan-1x3+C then the value of A+B is

I=∫x4+1x6+1dx=∫x4−x2+1+x2x6+1dx⇒ I=∫x4−x2+1x23+1dx+∫x2x6+1dx⇒ I=∫1x2+1dx+13∫1x32+1dx3⇒ I=tan−1⁡x+13tan−1⁡x3+C

∫x4+1x6+1dx=Atan-1x+1Btan-1x3+C then the value of A+B is

I=∫x4+1x6+1dx=∫x4−x2+1+x2x6+1dx⇒ I=∫x4−x2+1x23+1dx+∫x2x6+1dx⇒ I=∫1x2+1dx+13∫1x32+1dx3⇒ I=tan−1⁡x+13tan−1⁡x3+C

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$\int \frac{x^{4} + 1}{x^{6} + 1} dx = {Atan}^{- 1} x + \frac{1}{B} \tan^{- 1} (x^{3}) + C$ then the value of A+B is

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answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x^{4} + 1}{x^{6} + 1} d x = \int \frac{(x^{4} - x^{2} + 1) + x^{2}}{x^{6} + 1} d x \\ \Rightarrow I = \int \frac{x^{4} - x^{2} + 1}{{(x^{2})}^{3} + 1} d x + \int \frac{x^{2}}{x^{6} + 1} d x \\ \Rightarrow I = \int \frac{1}{x^{2} + 1} d x + \frac{1}{3} \int \frac{1}{{(x^{3})}^{2} + 1} d (x^{3}) \\ \Rightarrow I = \tan^{- 1} x + \frac{1}{3} \tan^{- 1} (x^{3}) + C \end{array}$