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Question Number 62855 by mathmax by abdo last updated on 26/Jun/19

f i n d \int {(\frac{x^{4}}{1 + x^{6}})}^{2} d x

2) c a l c u l a t e \int_{0}^{1} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x

3) c a l c u l a t e \int_{0}^{+ \infty} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x .

Commented by mathmax by abdo last updated on 27/Jun/19

1) l e t I = \int \frac{x^{8}}{{(1 + x^{6})}^{2}} d x \Rightarrow I = \int \frac{x^{8}}{{(1 + {(x^{3})}^{2})}^{2}} d x c h a n g e m e n t x^{3} = t a n θ g i v e

x = {(t a n θ)}^{\frac{1}{3}} \Rightarrow I = \int \frac{{(t a n θ)}^{\frac{8}{3}}}{{(1 + {t a n}^{2} θ)}^{2}} \frac{1}{3} (1 + {t a n}^{2} θ) {(t a n θ)}^{- \frac{2}{3}} d θ

= \frac{1}{3} \int \frac{t a \overset{2}{n} θ}{1 + {t a n}^{2} θ} d θ = \frac{1}{3} \int \frac{{s i n}^{2} θ}{{c o s}^{2} θ} {c o s}^{2} θ d θ = \frac{1}{3} \int {s i n}^{2} θ d θ = \frac{1}{3} \int \frac{1 - c o s (2 θ)}{2} d θ

= \frac{θ}{6} - \frac{1}{12} s i n (2 θ) + c b u t θ = a r c t a n (x^{3})

s i n (2 θ) = 2 s i n θ c o s θ = 2 s i n (a r c t a n (x^{3})) c o s (a r c t a n (x^{3}))

= 2 \frac{x^{3}}{\sqrt{1 + x^{6}}} . \frac{1}{\sqrt{1 + x^{6}}} = \frac{2 x^{3}}{1 + x^{6}} \Rightarrow I = \frac{1}{6} a r c t a n (x^{3}) - \frac{1}{6} \frac{x^{3}}{1 + x^{6}} + c

2) \int_{0}^{1} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x = \frac{1}{6} {[a r c t a n (x^{3}) - \frac{x^{3}}{1 + x^{6}}]}_{0}^{1} = \frac{1}{6} {\frac{π}{4} - \frac{1}{2}}

3) \int_{0}^{\infty} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x = \frac{1}{6} {[a r c t a n (x^{3}) - \frac{x^{3}}{1 + x^{6}}]}_{0}^{+ \infty} = \frac{1}{6} \frac{π}{2} = \frac{π}{12} .

\times