Evaluate ∫_(−∞) ^∞ 3x^2 (x^3 + 1)^2 e^(−x^6 − 2x^3 ) dx


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Question Number 33155 by Joel578 last updated on 11/Apr/18

$Evaluate$ $\int_{- \infty}^{\infty} 3 x^{2} {(x^{3} + 1)}^{2} e^{- x^{6} - 2 x^{3}} d x$
Commented by Joel578 last updated on 11/Apr/18

$My try$ $u = x^{3} + 1 \to d u = 3 x^{2} d x$ $I = \underset{- \infty}{\int^{\infty}} 3 x^{2} u^{2} e^{- x^{3} (u + 1)} (\frac{d u}{3 x^{2}})$ $= \underset{- \infty}{\int^{\infty}} u^{2} e^{- (u - 1) (u + 1)} d u$ $= \underset{- \infty}{\int^{\infty}} u^{2} e^{1 - u^{2}} d u$ $= e \underset{- \infty}{\int^{\infty}} u^{2} e^{- u^{2}} d u$ $= 2 e \underset{0}{\int^{\infty}} u^{2} e^{- u^{2}} d u$ $I dont know what to do next . . .$
Commented by prof Abdo imad last updated on 11/Apr/18

$l e t p u t I = \int_{- \infty}^{+ \infty} 3 x^{2} {(x^{3} + 1)}^{2} e^{- x^{6} - 2 x^{3}} d x . c h . x^{3} = t$ $g i v e I = \int_{- \infty}^{+ \infty} {(t + 1)}^{2} e^{- t^{2} - 2 t} d t$ $I = - \frac{1}{2} \int_{- \infty}^{+ \infty} (t + 1) (- 2 t - 2) e^{- t^{2} - 2 t} d t (b y p a r t s)$ $- 2 I = {[(t + 1) e^{- t^{2} - 2 t}]}_{- \infty}^{+ \infty} - \int_{- \infty}^{+ \infty} e^{- t^{2} - 2 t} d t \Rightarrow$ $2 I = \int_{- \infty}^{+ \infty} e^{- ({(t + 1)}^{2} - 1)} d t = e \int_{- \infty}^{+ \infty} e^{- {(t + 1)}^{2}} d t$ $=_{t + 1 = x} e \int_{- \infty}^{+ \infty} e^{- x^{2}} d x = e \sqrt{π} \Rightarrow I = \frac{e \sqrt{π}}{2}$ $b r c a u s e \int_{- \infty}^{+ \infty} e^{- x^{2}} d x = \sqrt{π} .$
Commented by Joel578 last updated on 13/Apr/18

$T h a n k y o u v e r y m u c h$
Commented by abdo imad last updated on 13/Apr/18

$n e v e r m i n d s i r j o e l .$
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