How do you integrate $(x^{5}) \cdot (e^{\frac{x^{2}}{2}})$ $(x^5)*(e^(x^2/2))$ ?

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How do you integrate $(x^{5}) \cdot (e^{\frac{x^{2}}{2}})$ $(x^5)*(e^(x^2/2))$ ?

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Answer 1 · 2015-09-01T15:09:48

$(4 x^{4} - 4 x^{2} - 8) e^{\frac{x^{2}}{2}} + c$ $(4x^4-4x^2-8)e^(x^2/2)+c$

Explanation:

Let's assume $\frac{x^{2}}{2} = t$ $x^2/2 = t$
$\Rightarrow \frac{2 x d x}{2} = t, x d x = d t$ $=> (2xdx)/2 = t, x dx = dt$

$= \int (x^{5}) \cdot (e^{\frac{x^{2}}{2}}) d x$ $=int(x^5)*(e^(x^2/2))dx$ can be written as,

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

Substituting with $t$ $t$ we get

$= \int (2 t) (2 t) (e^{t}) d t$ $=int(2t)(2t)(e^t)dt$

$= 4 \int t^{2} (e^{t}) d t$ $=4intt^2(e^t)dt$

Using Integration by Parts,

$∫(I)(II)dx=(I)∫(II)dx−∫((I)'∫(II)dx)dx$

where $(I)$ and $(II)$ are functions of $x$ , and $(I)$ represents which will be differentiated and $(II)$ will be integrated subsequently in the above formula

Similarly following for the problem,

$=4(t^2)inte^tdt-4int((t^2)'inte^tdt)dt+c$

$=4t^2e^t-4int2te^tdt+c$

$=4t^2e^t-8intte^tdt+c$

Applying again integration by parts in second term, we get

$=4t^2e^t-8(te^t-inte^tdt)+c$

$=4t^2e^t-8(te^t-e^t)+c$

$=4t^2e^t-8te^t-8e^t+c$

Substituting $t$ back,

$=4(x^2/2)^2e^(x^2/2)-8(x^2/2)e^(x^2/2)-8e^(x^2/2)+c$ , where $c$ is a constant

$=(4x^4-4x^2-8)e^(x^2/2)+c$

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How do you integrate $(x^{5}) \cdot (e^{\frac{x^{2}}{2}})$ $(x^5)*(e^(x^2/2))$ ?

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How do you integrate $(x^{5}) \cdot (e^{\frac{x^{2}}{2}})$ $(x^5)*(e^(x^2/2))$ ?