$\displaystyle \int 3xe^{2x^2} \ \mathrm{d}x$
How do you integrate this? Thanks in advance.
Hello,
By the substitution : $\displaystyle t=2x^2$
How can you know it's this way ? Because you can notice that the derivative of $\displaystyle 2x^2$ is $\displaystyle 4x$.
In front of the exponential, we have $\displaystyle 3x=\frac 34 (4x)$
So you have something like $\displaystyle k \int u'(x)e^{u(x)} dx$, k is a constant (here $\displaystyle \frac 34$)
Here, two methods :
- you substitute v=u(x)
- you recognize the derivative of $\displaystyle e^{u(x)}$