Question about integration by parts

Question

Could someone please explain how we calculated the first term of the second identity?

Thanks a ton.

Answer 1

Using integration by part with $f'(x)=1$ and $g(x)=\frac{1}{x^2+1}$ yields $$ \int\frac{1}{x^2+1}\,dx =x\cdot\frac{1}{x^2+1}-\int x\cdot\frac{-2x}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\frac{x^2}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\frac{(x^2+1)-1}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\left(\frac{1}{x^2+1}-\frac{1}{(x^2+1)^2}\right)\,dx\\ =\frac{x}{x^2+1}+2\int\frac{1}{x^2+1}\,dx-2\int\frac{1}{(x^2+1)^2}\,dx $$

For the second equality set $$ I=\int\frac{1}{(x^2+1)^2}\,dx\qquad J=\int\frac{1}{x^2+1}\,dx $$ In part 1 we proved that $$ J=\frac{x}{x^2+1}+2J-2I $$ Hence $$ 2I=\frac{x}{x^2+1}+J $$ that is $$ I=\frac{x}{2(x^2+1)}+\frac{J}{2} $$ as requared.