antidifferentiate : 2xe^(-x^2)

Question

anonymous · Answer

$$\int\limits_{}^{}2xe ^{-x ^{2}}$$

john_es · Answer

$$-e^{x^2}$$

anonymous · Answer

how do u do it?

john_es · Answer

Well, it is an inmediate integral, because
$$\int g'(x)f(g(x))dx$$

anonymous · Answer

$$\int\limits_{}^{}2xe ^{-x ^{2}} = -\int e^{-x^2}d(-x^2)$$

john_es · Answer

But you can try with a change of variable,
$$x^2=t\2xdx=dt$$

anonymous · Answer

so what happens if its g'x f(g(x)?

john_es · Answer

In this particular case, we have
$$\int g'(x)f'(g(x))dx=\int \frac{d(f(g(x)))}{dx} dx=f(g(x))$$

john_es · Answer

When you have integrals that cannot solve this way, but still are "inmediate", you can do a change of variable. I suggested x^2=t, but watching the answer of @wio you can also do, -x^2=t.
$$-x^2=t\
-2xdx=dt\Rightarrow2xdx=-dt\
\int 2xe^{x^2}dx=-\int e^tdt=-e^t=-e^{x^2}$$Don't forget to add the usual constant for indefinite integrals,
$$-e^{x^2}+K$$

anonymous · Answer

Thankyou

john_es · Answer

;)