Answer
$-2\ln|x+1|+\ln(x^{2}+1)+2\tan^{-1}x+C$
Work Step by Step
$I=\displaystyle \int\frac{4x}{x^{3}+x^{2}+x+1}dx$
$x^{3}+x^{2}+x+1=x^{2}(x+1)+(x+1)=(x+1)(x^{2}+1)$
$\displaystyle \frac{4x}{(x+1)(x^{2}+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}$
$4x=A(x^{2}+1)+(Bx+C)(x+1)$
$4x=Ax^{2}+A+Bx^{2}+Bx+Cx+C$
$4x=x^{2}(A+B)+x(B+C)+(A+C)$
$A+B=0\Rightarrow A=-B$
$A+C=0\Rightarrow C=-A=B$
$ B+C=4\Rightarrow 2B=4\Rightarrow B=C=2,A=-2$
$\displaystyle \frac{4x}{(x+1)(x^{2}+1)}=\frac{-2}{x+1}+\frac{2x+2}{x^{2}+1}=\frac{-2}{x+1}+\frac{2x}{x^{2}+1}+\frac{2}{x^{2}+1}$
$I=-2\ln|x+1|+\ln(x^{2}+1)+2\tan^{-1}x+C$
