find ∫ ((x^4 /(1+x^6 )))^2 dx 2) calculate ∫_0 ^1 (x^8 /((1+x^6 )^2 ))dx 3) calculate ∫


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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
$1) let I =∫ (x^8 /((1+x^6 )^2 ))dx ⇒I =∫ (x^8 /((1+(x^3 )^2 )^2 ))dx changement x^3 =tanθ give x =(tanθ)^(1/(3 )) ⇒ I =∫ (((tanθ)^(8/3) )/((1+tan^2 θ)^2 )) (1/3)(1+tan^2 θ)(tanθ)^(−(2/3)) dθ =(1/3) ∫ ((tan^2 θ)/(1+tan^2 θ)) dθ =(1/3) ∫ ((sin^2 θ)/(cos^2 θ)) cos^2 θ dθ =(1/3) ∫ sin^2 θ dθ =(1/3) ∫((1−cos(2θ))/2)dθ =(θ/6) −(1/(12)) sin(2θ) +c but θ =arctan(x^3 ) sin(2θ) =2sinθ cosθ =2 sin(arctan(x^3 ))cos(arctan(x^3 )) =2 (x^3 /(√(1+x^6 ))) .(1/(√(1+x^6 ))) =((2x^3 )/(1+x^6 )) ⇒ I =(1/6) arctan(x^3 )−(1/6)(x^3 /(1+x^6 )) +c 2) ∫_0 ^1 (x^8 /((1+x^6 )^2 ))dx =(1/6)[arctan(x^3 )−(x^3 /(1+x^6 ))]_0 ^1 =(1/6){(π/4) −(1/2)} 3) ∫_0 ^∞ (x^8 /((1+x^6 )^2 ))dx =(1/6)[ arctan(x^3 )−(x^3 /(1+x^6 ))]_0 ^(+∞) =(1/6) (π/2) =(π/(12)) . ×$
$1) l e t I = \int \frac{x^{8}}{{(1 + x^{6})}^{2}} d x \Rightarrow I = \int \frac{x^{8}}{{(1 + {(x^{3})}^{2})}^{2}} d x c h a n g e m e n t x^{3} = t a n θ g i v e$ $x = {(t a n θ)}^{\frac{1}{3}} \Rightarrow I = \int \frac{{(t a n θ)}^{\frac{8}{3}}}{{(1 + {t a n}^{2} θ)}^{2}} \frac{1}{3} (1 + {t a n}^{2} θ) {(t a n θ)}^{- \frac{2}{3}} d θ$ $= \frac{1}{3} \int \frac{t a \overset{2}{n} θ}{1 + {t a n}^{2} θ} d θ = \frac{1}{3} \int \frac{{s i n}^{2} θ}{{c o s}^{2} θ} {c o s}^{2} θ d θ = \frac{1}{3} \int {s i n}^{2} θ d θ = \frac{1}{3} \int \frac{1 - c o s (2 θ)}{2} d θ$ $= \frac{θ}{6} - \frac{1}{12} s i n (2 θ) + c b u t θ = a r c t a n (x^{3})$ $s i n (2 θ) = 2 s i n θ c o s θ = 2 s i n (a r c t a n (x^{3})) c o s (a r c t a n (x^{3}))$ $= 2 \frac{x^{3}}{\sqrt{1 + x^{6}}} . \frac{1}{\sqrt{1 + x^{6}}} = \frac{2 x^{3}}{1 + x^{6}} \Rightarrow I = \frac{1}{6} a r c t a n (x^{3}) - \frac{1}{6} \frac{x^{3}}{1 + x^{6}} + c$ $2) \int_{0}^{1} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x = \frac{1}{6} {[a r c t a n (x^{3}) - \frac{x^{3}}{1 + x^{6}}]}_{0}^{1} = \frac{1}{6} {\frac{π}{4} - \frac{1}{2}}$ $3) \int_{0}^{\infty} \frac{x^{8}}{{(1 + x^{6})}^{2}} d x = \frac{1}{6} {[a r c t a n (x^{3}) - \frac{x^{3}}{1 + x^{6}}]}_{0}^{+ \infty} = \frac{1}{6} \frac{π}{2} = \frac{π}{12} .$ $\times$
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