Question:
Evaluate the following integral:
$\int \frac{1}{x\left(x^{4}+1\right)} d x$
Solution:
Let
$I=\int \frac{1}{x\left(x^{4}+1\right)} d x$
$\frac{1}{x\left(x^{4}+1\right)}=\frac{A}{x}+\frac{B x^{3}+C x^{2}+D x+E}{x^{4}+1}$
$I=A\left(x^{4}+1\right)+\left(B x^{3}+C x^{2}+D x+E\right)(x)$
Equating constants
$A=1$
Equating coefficients of $x^{4}$
$0=A+B$
$0=1+B$
$B=-1$
Equating coefficients of $x^{2}$
$D=0$
Equating coefficients of $x$
$E=0$
Thus,
$I=\int \frac{d x}{x}+\int-\frac{x^{2} d x}{x^{4}+1}$
$=\log |x|-\frac{1}{4} \log \left|x^{4}+1\right|+C$
$=\frac{4}{4} \log |x|-\frac{1}{4} \log \left|x^{4}+1\right|+C$
$=\frac{1}{4} \log \left|x^{4}\right|-\frac{1}{4} \log \left|x^{4}+1\right|+C$
$\frac{1}{4} \log \left|\frac{x^{4}}{x^{4}+1}\right|+C$