Answer
$\frac{1}{x}+ln|x^{4}+x^{3}|+C$
Work Step by Step
$\int\frac{4x^{2}+2x-1}{x^{3}+x^{2}}dx$
Use partial fraction decomposition to split up the integral.
$\frac{4x^{2}+2x-1}{x^{2}(x+1)}$
$\frac{4x^{2}+2x-1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$(\frac{4x^{2}+2x-1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1})(x^{2})(x+1)$
$4x^{2}+2x-1=Ax(x+1)+B(x+1)+Cx^{2}$
Substitute in the values for $x$ that would cancel out terms:
$x=0$ -> $4(0)^{2}+2(0)-1=A(0)(0+1)+B(0+1)+C(0)^{2}$
$B=-1$
$x=-1$ -> $4(-1)^{2}+2(-1)-1=A(-1)(-1+1)+B(-1+1)+C(-1)^{2}$
$4-2-1=C$
$C=1$
Now substitute in a random value for $x$:
$x=1$ -> $4(1)^{2}+2(1)-1=A(1)(1+1)+B(1+1)+C(1)^{2}$
$4+2-1=2A+2B+C$
Substitute $B=-1$ and $C=1$:
$5=2A+2(-1)+(1)$
$5=2A-1$
$A=3$
$\frac{4x^{2}+2x-1}{x^{2}(x+1)}=\frac{3}{x}-\frac{1}{x^{2}}+\frac{1}{x+1}$
$\int\frac{3}{x}-\frac{1}{x^{2}}+\frac{1}{x+1}dx$
$3\int\frac{1}{x}dx-\int(x^{-2})dx+\int\frac{1}{x+1}dx$
$let$ $u=x+1$
$du=dx$
$3ln|x|+\frac{1}{x}+\int\frac{1}{u}du$
$3ln|x|+\frac{1}{x}+ln|u|+C$
$3ln|x|+\frac{1}{x}+ln|x+1|+C$
$ln|x^{3}|+\frac{1}{x}+ln|x+1|+C$
$\frac{1}{x}+ln|x^{4}+x^{3}|+C$
