Problem source

Find constants $a,\,b,\,c$ and $d$ such that
$$\frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2}=
\frac{1}{ x^4+4}.$$
Show that
$$\int_0^1\frac {\d x}{ x^4+4}\;= \frac{1}{16} \ln 5
 +\frac{1}{8} \tan^{-1}2 .$$

Solution source

First notice that $(x^2+2+2x)(x^2+2-2x) = (x^2+2)^2-4x^2 = x^2+4$ and so

\begin{align*}
&& \frac{1}{x^4+4} &= \frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2} \\
\Rightarrow && 1 &= (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) \\
&&&= (a+c)x^3+(b-2a+d+2c)x^2+(2a-2b+2c+2d)x+2b+2d \\
\Rightarrow && 0 &= a+c \\
&& 2a &= b+d+2c \\
&& b &= a+c+d \\
&& \frac12 &= b+d \\
\Rightarrow && c &= -a \\
&& 2a &= \frac12+2(-a) \\
\Rightarrow && a &= \frac18, c = -\frac18 \\
&& b &= d \\
\Rightarrow && b &= \frac14, d = \frac14
\end{align*}

Therefore

\begin{align*}
\int_0^1\frac {\d x}{ x^4+4} &= \int_0^1 \frac18\left ( \frac{x+2}{x^2+2x+2} -\frac{x-2}{x^2-2x+2}\right) \d x \\
&= \frac1{16}\int_0^1 \left ( \frac{2x+2+2}{x^2+2x+2} -\frac{2x-2-2}{x^2-2x+2}\right) \d x \\
&= \frac1{16} \left [ \ln(x^2+2x+2) -\ln(x^2-2x+2)\right]_0^1 + \frac1{16} \int_0^1 \left ( \frac{2}{(x+1)^2+1} + \frac{2}{(x-1)^2+1} \right) \d x \\
&= \frac1{16} \left (\ln 5 - \ln 1 -(\ln 2-\ln 2) \right) + \frac18 \left [ \tan^{-1} (x+1)+\tan^{-1}(x-1) \right]_0^1 \\
&= \frac1{16} \ln 5 + \frac18 \left (\tan^{-1} 2+\tan^{-1} 0 - \tan^{-1}1-\tan^{-1} (-1) \right) \\
&=  \frac1{16} \ln 5+ \frac18 \tan^{-1} 2
\end{align*}