`intt^3e^(-t^2)dt`

Let `x=t^2`

`dx=2tdt`

`intt^3e^(-t^2)dt=intxe^(-x)dx/2`

`=1/2intxe^(-x)dx`

Now apply integration by parts,

If f(x) and g(x) are differentiable functions then,

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int((du)/dxintvdx)dx`

So, let's take u=x , then u'=1

and v=`e^-x`

then v'=`-e^-x`

`intxe^-xdx=x*int(e^-xdx)-int(1inte^-xdx)dx`

`=x(-e^-x)-int(-e^-x)dx`

`=-xe^-x+int(e^-x)dx`

`=-xe^-x+(-e^-x)`

`=-xe^-x-e^-x`

`:.intt^3e^(-t^2)dt=1/2(-xe^-x-e^-x)`

substitute back `x=t^2` and add a constant to the solution,

`intt^3e^(-t^2)dt=1/2(-t^2e^(-t^2)-e^(-t^2))+C`

` `

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