Question

Show that $\int \frac{x-1}{(x+2)(x^2+1)}dx=\log (x+1)+\frac{1}{2}\log (x^2+1)$.

Answer 1

$I=\int \frac{x-1}{(x+2)(x^2+1)}dx=\int (\frac{3x-1}{5(x^2+1)}-\frac{3}{5(x+2)})dx$
$=\frac{1}{5}\int \frac{3x-1}{x^2+1}dx-\frac{3}{5}\int \frac{1}{x+2}dx$
$=\frac{1}{5}\int (\frac{3x}{x^2+1}-\frac{1}{x^2+1})dx-\frac{3}{5}\int \frac{1}{x+2}dx$
$=\frac{3}{5}\int \frac{x}{x^2+1}dx-\frac{1}{5}\int \frac{1}{x^2+1}dx-\frac{3}{5}\int \frac{1}{x+2}dx$
$=\frac{3}{10}\log (x^2+1)-\frac{1}{5}{\tan }^{-1}x-\frac{3}{5}\log (x+2)$