Answer
$\ln|x|-5\tan^{-1}x+C$
Work Step by Step
$\int (\frac{1}{x}-\frac{5}{x^{2}+1})dx=$$\int \frac{1}{x}dx-5\int\frac{dx}{x^{2}+1^2}=$
$\ln|x|-5\times\frac{1}{1}\tan^{-1}\frac{x}{1}+C$ (As $\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a})$
$=\ln|x|-5\tan^{-1}x+C$
