∫ x 4 + 1 1 + x 6 d x = Options: A . &nbsptan−1(x)−tan−1(x3)+c B . &nbsptan−1(x)−13tan−1(x3)+c C . &nbsptan−1(x)+tan−1(x3)+c D . &nbsptan−1(x)+13tan−1(x3)+c

Answer: Option D : D I = ∫ x 4 + 1 1 + x 6 d x = ∫ ( x 4 − x 2 + 1 ) + x 2 ( 1 + x 6 ) d x = ∫ x 4 − x 2 + 1 1 + x 6 d x + ∫ x 2 1 + x 6 d x = ∫ 1 1 + x 2 d x + 1 3 ∫ 3 x 2 1 + x 6 d x = t a n − 1 ( x ) + 1 3 t a n − 1 x 3 + c

∫x4+11+x6 dx=

Question

\int \frac{x^{4} + 1}{1 + x^{6}} d x =

Options:

A . tan−1(x)−tan−1(x3)+c

B . tan−1(x)−13tan−1(x3)+c

C . tan−1(x)+tan−1(x3)+c

D . tan−1(x)+13tan−1(x3)+c

Answer: Option D
:
D

I = \int \frac{x^{4} + 1}{1 + x^{6}} d x = \int \frac{(x^{4} - x^{2} + 1) + x^{2}}{(1 + x^{6})} d x = \int \frac{x^{4} - x^{2} + 1}{1 + x^{6}} d x + \int \frac{x^{2}}{1 + x^{6}} d x = \int \frac{1}{1 + x^{2}} d x + \frac{1}{3} \int \frac{3 x^{2}}{1 + x^{6}} d x = t a n^{- 1} (x) + \frac{1}{3} t a n^{- 1} x^{3} + c

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