Answer
The solution is $$\int\frac{1}{x^2(x+1)}dx=-\ln|x|-\frac{1}{x}+\ln|x+1|+c.$$
Work Step by Step
To solve the integral $$\int\frac{1}{x^2(x+1)}dx$$ we will firstly transform the function under the integral using partial fractions. This function can be decomposed as:
$$\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1},$$
where $A,B,C$ are constants that we need to find. We will find them following the steps below.
Step 1: We will put the partial fractions together:
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}=\frac{Ax(x+1)+B(x+1)+Cx^2}{x^2(x+1)}=\frac{Ax^2+Ax+Bx+B+Cx^2}{x^2(x+1)},$$
so we have
$$ \frac{1}{x^2(x+1)}=\frac{Ax^2+Ax+Bx+B+Cx^2}{x^2(x+1)}.$$
Step 2: Now we will equate the numerators on the both sides:
$$Ax^2+Ax+Bx+B+Cx^2=1.$$
The coefficients multiplying powers of $x$ have to match on both sides, so we have: $B=1,A+B=0, A+C=0$. Putting $B=1$ into second equation we get $A=-1$, and then putting $A=-1$ into the third equation we have $C=1$.
Now we return all this into original integral:
$$\int\frac{1}{x^2(x+1)}dx=-\int\frac{dx}{x}+\int\frac{dx}{x^2}+\int\frac{dx}{x+1}$$
Let's solve those integrals:
Integral 1: $$\int\frac{dx}{x}=\ln|x|+c_1.$$
Integral 2: $$\int\frac{dx}{x^2}=-\frac{1}{x}+c_2.$$
Integral 3: By using substitution $t=x+1$ we have $dt=dx$ and putting this into Integral 3 we have
$$\int\frac{dx}{x+1}=\int\frac{dt}{t}=\ln|t|+c_3$$.
Now we have to express $t$ in terms of $x$ and then we have
$$\int\frac{dx}{x+1}=\ln|x+1|+c_3.$$
Constants $c_1,c_2,c_3$ are arbitrary, so putting $c=c_1+c_2+c_3$ it will be arbitrary as well.
Finally, putting the solutions for integral 1, 2 and 3 into our original integral we have
$$\int\frac{1}{x^2(x+1)}dx=-\ln|x|-\frac{1}{x}+\ln|x+1|+c.$$
