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

1 Answer

\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