There is the integral: $$\int e^{-2x}\left(\frac{1}{x^2}+\frac{3}{x}\right)dx$$
The integral cannot be calculated if divide it into sum of two integrals and try to solve them one by one. But there is the way to solve it as a whole sum.
The way exists because of that example if we assume that $f(x)$ is function with non-computable integral $\int(x-f(x))dx$ is non-computable too. But $\int f(x)dx + \int (x-f(x))dx$ can be calculated.
$\int e^{-2x}\left(\dfrac{1}{x^2}+\dfrac{3}{x}\right)~dx$
$=\int\dfrac{e^{-2x}}{x^2}~dx+\int\dfrac{3e^{-2x}}{x}~dx$
$=-\int e^{-2x}~d\left(\dfrac{1}{x}\right)+\int\dfrac{3e^{-2x}}{x}~dx$
$=-\dfrac{e^{-2x}}{x}+\int\dfrac{1}{x}~d\left(e^{-2x}\right)+\int\dfrac{3e^{-2x}}{x}~dx$
$=-\dfrac{e^{-2x}}{x}-\int\dfrac{2e^{-2x}}{x}~dx+\int\dfrac{3e^{-2x}}{x}~dx$
$=-\dfrac{e^{-2x}}{x}+\int\dfrac{e^{-2x}}{x}~dx$
$=-\dfrac{e^{-2x}}{x}+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^nx^{n-1}}{n!}~dx$
$=-\dfrac{e^{-2x}}{x}+\int\left(\dfrac{1}{x}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^nx^{n-1}}{n!}\right)~dx$
$=-\dfrac{e^{-2x}}{x}+\ln x+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^nx^n}{n!n}+C$