How do you integrate int x^3*e^(x^2/2) ∫x3⋅ex22 using integration by parts?

2 Answers
Cem Sentin
Nov 16, 2017

x^3*e^[x^2/2]*dx=(x^2-2)*e^(x^2/2)+Cx3⋅ex22⋅dx=(x2−2)⋅ex22+C

Explanation:

x^3*e^[x^2/2]*dxx3⋅ex22⋅dx

=x^2*xe^(x^2/2)*dxx2⋅xex22⋅dx

I used integration by parts with u=x^2u=x2 and dv=xe^[(x^2/2)]*dxdv=xe(x22)⋅dx, so du=2x*dxdu=2x⋅dx and v=e^(x^2/2)v=ex22

u*dv=u*v-int v*duu⋅dv=u⋅v−∫v⋅du

=x^2*xe^[(x^2/2)]*dxx2⋅xe(x22)⋅dx

=x^2*e^(x^2/2)-int 2x*e^(x^2/2)*dxx2⋅ex22−∫2x⋅ex22⋅dx

=x^2*e^(x^2/2)-2e^(x^2/2)+Cx2⋅ex22−2ex22+C

=(x^2-2)*e^(x^2/2)+C(x2−2)⋅ex22+C

Narad T.
Nov 16, 2017

The answer is =(x^2-2)e^(x^2/2)+C=(x2−2)ex22+C

Explanation:

We need

intpq'=pq-intp'q

Here,

Perform the substitution

u=x^2, =>, du=2xdx, =>, dx=1/(2x)du

So,

intx^3e^(x^2/2)dx=1/2intue^(u/2)du

Perform the integration by parts

p=u, =>, p'=1

q'=e^(u/2), =>, q=2e^(u/2)

Therefore,

intx^3e^(x^2/2)dx=1/2intue^(u/2)du

=1/2(2ue^(u/2)-2inte^(u/2)du)

=1/2(2ue^(u/2)-4e^(u/2))

=x^2e^(x^2/2)-2e^(x^2/2)+C

=(x^2-2)e^(x^2/2)+C

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