Answer
$$\int\frac{x^2dx}{(x-1)(x^2+2x+1)}=\frac{1}{4}\ln|x-1|+\frac{3}{4}\ln|x+1|+\frac{1}{2(x+1)}+C$$
Work Step by Step
$$A=\int\frac{x^2dx}{(x-1)(x^2+2x+1)}=\int\frac{x^2dx}{(x-1)(x+1)^2}$$
1) Express the integrand as a sum of partial fractions:
$$\frac{1}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$
Clear fractions:
$$A(x+1)^2+B(x^2-1)+C(x-1)=x^2$$ $$Ax^2+2Ax+A+Bx^2-B+Cx-C=x^2$$ $$(A+B)x^2+(2A+C)x+(A-B-C)=x^2$$
Equating coefficients of corresponding powers of $x$, we get
- $A+B=1$
- $2A+C=0$
- $A-B-C=0$
Solving for $A$, $B$, $C$, we get $A=1/4$, $B=3/4$ and $C=-1/2$.
Therefore,
$$\frac{1}{(x-1)(x+1)^2}=\frac{1}{4}\frac{1}{x-1}+\frac{3}{4}\frac{1}{x+1}-\frac{1}{2}\frac{1}{(x+1)^2}$$
2) Evaluate the integral:
$$A=\frac{1}{4}\int\frac{dx}{x-1}+\frac{3}{4}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{(x+1)^2}$$ $$A=\frac{1}{4}\ln|x-1|+\frac{3}{4}\ln|x+1|+\frac{1}{2(x+1)}+C$$
