Evaluate the Integral: $\int(x^5+5^x)\ dx$ $\int(x^5+5^x)\ dx$
I made the the terms within the parenthesis u
$u=x^5+5^x$
$du=5x^4+5^xln\ 5$ 
$du=5x^4+5^x\ ln\ 5 dx$
$\frac{u}{5x^4+5^xln\ 5}\ du$
I am stuck at this point. Is there a better way to attack this problem? I know I have to use this formula 

$\int\ a^x dx=\frac{a^x}{ln\ a}+C$

But how? 
 A: You are making it too complicated. You just need to use the rules
$$
\int x^n \; dx = \frac{1}{n+1}x^{n+1} + C\\
$$
and
$$
\int a^x \; dx= \frac{1}{\ln(a)}a^x + C.
$$
So
$$
\int x^5 + 5^x \; dx = \int x^5 \; dx + \int 5^x \; dx = ...
$$
A: Hint:
You integral is a sum:
$$
\int (x^5+5^x)dx=\int x^5 dx+\int 5^x dx
$$
now use $5=e^{log 5}$ and you have:
$$
\int x^5 dx+\int e^{xlog 5} dx
$$
can you integrate this?
A: Notice, $$\int (x^5+5^x)dx$$ $$=\int x^5 dx+\int 5^x dx$$ $$=\frac{x^6}{6}+\frac{5^x}{\ln 5}+C$$
A: Split the integral into two: $$\int x^5 \mathrm {d}x +\int5^x \mathrm {d}x$$ and evaluate them.$$\int x^5 \mathrm {d}x=\frac{x^6}{6}$$ and $$\int5^x \mathrm {d}x=\frac{5^x}{\mathrm {ln} 5}$$ then add them together to get the answer: $$\int x^5 \mathrm {d}x +\int5^x \mathrm {d}x=\frac{x^6}{6}+\frac{5^x}{\mathrm {ln} 5}+C$$
A: If you want to evaluate this integral, you should simplify like this:
$\int{(x^5 + 5^x) dx} = \int{x^5 dx} + \int{5^x dx}$. You can always split up an expression of the same variable like this to simplify your computation. Now, generally, $\int{x^ndx} = \frac{x^{n+1}}{n+1} +C$, and $\int{a^xdx} = \frac{a^x}{ln{(a)}} +C$.
So, our result is: $\int{x^5 dx} + \int{5^x dx} = \frac{x^6}{6} + \frac{5^x}{ln{(5)}} +C$. $C$ is the constant of integration, due to the integral being indefinite.
