Integration by parts of a function

Question

I have the following function to integrate by parts can't go around that on this assignment, i can't figure out what is the f and g and their derivatives respectively.

$$ ∫x^3e^{-x^2}dx $$

Thanks in advance for the help!

Answer 1

$$u=x^2, dv=xe^{-x^2}dx, v=-\frac{1}{2}e^{-x^2}, du=2xdx$$

so

$$\int x^3e^{-x^2}dx=-\frac{1}{2}x^2e^{-x^2}+\int xe^{-x^2}dx$$

where the new integral can be easily calculated by substitution $u=x^2$.

Answer 2

Hint:

Exponentials of functions are quite often impossible to integrate, unless multipied by the derivative of the function to form an expression like

$$f'(x)e^{f(x)}.$$

When you see such an expression, jump on it !

Here you observe

$$x^3e^{-x^2}=-\frac{x^2}2(-x^2)'e^{-x^2}.$$

The rest is yours.

Answer 3

hint

Put $$u (x)=-\frac {x^2}{2}$$ and $$v'(x)=-2xe^{-x^2} $$

thus $$u'(x)=-x $$ and $$v (x)=e^{-x^2} $$

it becomes $$\Bigl [u (x)v(x)\Bigr]+\int xe^{-x^2}dx $$

$$=u (x)v (x)-\frac{1}{2}v (x) $$ $$=-\frac {x^2+1}{2}e^{-x^2} +C $$