Question

How do you evaluate the integral `int x^3e^(x^2)"d"x`?

Answer 1

`\int x^3e^{x^2}\ dx=1/2(x^2-1)e^{x^2}+C`

Explanation:

Let `x^2=t\implies 2x=dt\ or \ xdx=\frac{dt}{2}`
`\therefore \int x^3e^{x^2}\ dx`
`=\int x^2e^{x^2}(xdx)`
`=\int te^t\frac{dt}{2}`
`=\frac{1}{2}\int te^t\ dt`
`=\1/2(t\int e^t \dt-\int (\frac{d}{dt}(t)\cdot \int e^t\ dt)dt)`
`=\1/2(te^t-\int (1\cdot e^t)dt)`
`=\1/2(te^t-\int e^tdt)`
`=\1/2(te^t-e^t)+C`
`=\1/2(t-1)e^t+C`
`=1/2(x^2-1)e^{x^2}+C`