I already know how to begin because I got the hint from an online calculator. The only thing is I can't derive the hint myself.
The hint is to write
$$5^x= \frac {\ln3}{\ln3-\ln5}(5^x+3^x)- \frac{1}{\ln3-\ln5}(\ln5 \cdot 5^x+ \ln3\cdot 3^x) $$
I don't have a clue on how to arrive at this hint. I know however how to continue after the hint.
Could some explain how one arrives at the hint?
I'm not a big fan of that approach. I would approach the problem as follows:
To begin, divide the top and bottom by $3^x$ to get $$ \int \frac{1}{1 + (5/3)^x}\,(5/3)^x dx $$ Now, it suffices to apply a $u$-subsitution with $u = (5/3)^x$. The integral is rewritten as $$ \frac 1{\ln(5/3)} \int \frac{1}{1 + u}\,du = \frac 1{\ln5 - \ln 3} \int \frac{1}{1 + u}\,du $$
As for how one arrives at the hint, I assume the thought process is as follows:
We would like to find coefficients $\alpha, \beta$ such that $$ 5^x = \alpha\,(5^x + 3^x) + \beta\,(\ln 5 \cdot 5^x + \ln 3\cdot 3^x) $$ once we have computed such coefficients, we can express the integral as $$ \int \frac{\alpha\,(5^x + 3^x) + \beta\,(\ln 5 \cdot 5^x + \ln 3\cdot 3^x)}{5^x + 3^x} dx =\\ \alpha\,x + \beta \ln|5^x + 3^x| + C $$ With that in mind: we can set up the equation for these coefficients as $$ 5^x = \alpha\,(5^x + 3^x) + \beta\,(\ln 5 \cdot 5^x + \ln 3\cdot 3^x) \implies\\ 5^x = (\alpha + \beta \ln 5) 5^x + (\alpha + \beta \ln 3) 3^x $$ Setting coefficients equal, we can now solve the system of equations $$ \alpha + \beta \ln 5 = 1\\ \alpha + \beta \ln 3 = 0 $$ to obtain the coefficients.
I'm sure there's a nice analogy to be made between this approach and the use of "partial fractions" for integrating rational expressions.