Help with integrating two base $e$ exponentials multiplied together. I was wondering if anyone could help me with integrating two base $e$ exponentials. I'm not really looking for help with a specific version of this but just help with this form of integral in general.
$$\int\limits_0^{10}(e^x-4)(e^{5x})\,dx$$
I'm guessing this can't be done using integration by parts since $\int e^x\,dx = e^x$.
I saw some videos suggesting $u$-substitution but I don't know if that would be the best way to approach something like this.
Any help  is appreciated.
 A: For your particular integral it's helpful to multiply it out first:
$$\int_0^{10}(e^x-4)(e^{5x})\,dx = \int_0^{10}\left(e^{6x} - 4e^{5x}\right)\,dx.$$
Now, if you have some experience with integration, you can recognize the antiderivative:
$$\int_0^{10}\left(e^{6x} - 4e^{5x}\right)\,dx = \left(\frac{1}{6}e^{6x} - \frac{4}{5}e^{5x}\right)\Bigg|_{0}^{10} = \frac{1}{6}e^{60}-\frac{4}{5}e^{50}- \frac{1}{6} + \frac{4}{5}\approx 1.9029 \times 10^{25}.$$
If that's too much of a jump, we can split the integral and use $u$ substitution:
$$\int_0^{10}\left(e^{6x} - 4e^{5x}\right)\,dx = \int_{0}^{10}e^{6x}\,dx - 4\int_{0}^{10}e^{5x}\,dx.$$
For the first integral, we can use $u = 6x$ and $du = 6\,dx$ to get
$$\int_{0}^{10}e^{6x}\,dx = \frac{1}{6}\int_{0}^{60}e^{u}\,du = \frac{1}{6}e^{u}\bigg|_{0}^{60} = \frac{1}{6}e^{60} - \frac{1}{6}.$$
A similar substitution will work to get the second integral.
In general, just because an integral is presented as a product doesn't mean integration by parts is going to be the best approach.  Sometimes it really is best to multiply it out first.
A: For exponentials of the same base you have the rule:
$$e^a\times e^b=e^{a+b}$$
and this can be extended to any base as I said:
$$c^a\times c^b=c^{a+b}$$

In your case this means:
$$(e^x-4)e^{5x}=e^xe^{5x}-4e^{5x}=e^{6x}-4e^{5x}$$
Now your integral becomes:
$$\int_0^{10}\left[e^{6x}-4e^{5x}\right]\,dx=\left[\frac{e^{6x}}{6}-\frac{4e^{5x}}{5}\right]_{x=0}^{10}$$
$$=\frac{e^{60}}{6}-\frac{4e^{50}}{5}-\frac16+\frac45$$
now you may wish to factorise this for the sake of calculation, so:
$$e^{50}\left(\frac{e^{10}}6-\frac45\right)+\frac{19}{30}$$
