The Notes: `int_0^1 (dx)/(x^2 + 1)^2` Evaluate the integral

Sunday, December 22, 2013

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}$ Evaluate the integral

First find the indefinite integral:

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}=\int\frac{{{1}+{{x}}^{{2}}-{{x}}^{{2}}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}{\left.{d}{x}\right.}=\int\frac{{{\left.{d}{x}\right.}}}{{{1}+{{x}}^{{2}}}}-\int\frac{{{{x}}^{{2}}{\left.{d}{x}\right.}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}.$

The first summand is $\displaystyle{\arctan{{\left({x}\right)}}},$ for the second apply integration by parts:

$\displaystyle{u}={x},$ $\displaystyle{d}{v}=\frac{{{x}{\left.{d}{x}\right.}}}{{{1}+{{x}}^{{2}}}},$ so $\displaystyle{d}{u}={\left.{d}{x}\right.}$ and $\displaystyle{v}=-\frac{{1}}{{2}}\frac{{1}}{{{1}+{{x}}^{{2}}}}.$

So $\displaystyle\int\frac{{{{x}}^{{2}}{\left.{d}{x}\right.}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}=-\frac{{1}}{{2}}\frac{{x}}{{{1}+{{x}}^{{2}}}}+\frac{{1}}{{2}}\int\frac{{{\left.{d}{x}\right.}}}{{{1}+{{x}}^{{2}}}}=-\frac{{1}}{{2}}\frac{{x}}{{{1}+{{x}}^{{2}}}}+\frac{{1}}{{2}}{\arctan{{\left({x}\right)}}}.$