Answer
$ \dfrac{-5}{x}+2 \tan^{-1} x+c$
Work Step by Step
Calculate the anti-derivative.
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $ \int \dfrac{1}{1+x^2} dx=tan^{-1} x+c$
where $c$ is a constant of proportionality.
Then $\int (\dfrac{5}{x^2}+\dfrac{2}{x^2+1}) dx=5 \int x^{-2} dx+2 \tan^{-1} x +c$
or, $= \dfrac{5x^{-2+1}}{-2+1}+2 \tan^{-1} x+c$
Thus, $= \dfrac{-5}{x}+2 \tan^{-1} x+c$
