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