How do you find the integral of $x^{5} \cdot e^{x^{2}}$ $x^5*e^(x^2)$ ?

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Answer 1 · 2018-04-25T10:50:16

$\frac{1}{2} x^{4} e^{x^{2}} - x^{2} e^{x^{2}} + e^{x^{2}} + C$ $1/2x^4e^(x^2)-x^2e^(x^2)+e^(x^2)+C$

$\int x^{5} \cdot e^{x^{2}} d x$ $intx^5*e^(x^2)dx$

$= \frac{1}{2} \int x^{4} e^{x^{2}} \cdot 2 x d x$ $=1/2intx^4e^(x^2)*2xdx$

$= \frac{1}{2} \int x^{4} e^{x^{2}} {d x}^{2}$ $=1/2intx^4e^(x^2)dx^2$

$x^{2} = u$ $x^2=u$

$d (x^{2}) = d u$ $d(x^2)=du$

$= \frac{1}{2} \int x^{4} e^{x^{2}} {d x}^{2} = \frac{1}{2} \int u^{2} e^{u} d u$ $=1/2intx^4e^(x^2)dx^2=1/2intu^2e^udu$

$= \frac{1}{2} \int u^{2} d (e^{u})$ $=1/2intu^2d(e^u)$

$= \frac{1}{2} (u^{2} e^{u} - \int 2 u e^{u} d u)$ $=1/2(u^2e^u-int2ue^udu)$

$= \frac{1}{2} (u^{2} e^{u} - 2 \int u d (e^{u}))$ $=1/2(u^2e^u-2intud(e^u))$

$= \frac{1}{2} (u^{2} e^{u} - 2 u e^{u} - (- 2 \int e^{u} d u))$ $=1/2(u^2e^u-2ue^u-(-2inte^udu))$

$= \frac{1}{2} u^{2} e^{u} - u e^{u} + e^{u} + C$ $=1/2u^2e^u-ue^u+e^u+C$

Reverse The Substitution

$= \frac{1}{2} x^{4} e^{x^{2}} - x^{2} e^{x^{2}} + e^{x^{2}} + C$ $=1/2x^4e^(x^2)-x^2e^(x^2)+e^(x^2)+C$

How do you find the integral of x5⋅ex2x^5*e^(x^2) ?