How to integrate 1 over x^2(x+1) using partial fractions

Expert intro: This is a classic partial-fractions integral. The main thing is to split the denominator carefully, then integrate each simple term without rushing the algebra.

Detailed walkthrough

Let

$\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}.$

Multiply both sides by $x^2(x+1)$:

$1=A x(x+1)+B(x+1)+Cx^2.$

Expand:

$1=Ax^2+Ax+Bx+B+Cx^2.$

Collect like terms:

$1=(A+C)x^2+(A+B)x+B.$

Now match coefficients with the constant polynomial $1$:

• $A+C=0$
• $A+B=0$
• $B=1$

So

$B=1,\quad A=-1,\quad C=1.$

Thus

$\frac{1}{x^2(x+1)}=-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}.$

Integrate term by term:

$\int \frac{dx}{x^2(x+1)}=\int\left(-\frac1x+\frac1{x^2}+\frac1{x+1}\right)dx.$

So

$= -\ln|x| - \frac{1}{x} + \ln|x+1| + C.$

A neat final form is

$\boxed{\ln\left|\frac{x+1}{x}\right| - \frac{1}{x} + C}.$

💡 Pitfall guide

A common slip is to write the partial fraction as $\frac{Ax+B}{x^2}+\frac{C}{x+1}$ and then forget that $\frac{Ax+B}{x^2}=\frac{A}{x}+\frac{B}{x^2}$. That still works, but the coefficient matching is easier if you expand into $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$ straight away. Also remember the integral of $x^{-2}$ is $-x^{-1}$, not $\ln x$.

🔄 Real-world variant

If the denominator were $x^2(x-a)$ instead of $x^2(x+1)$, the same method would work with

$\frac{1}{x^2(x-a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-a}.$

The coefficients would change with $a$, but the structure of the answer would still be a mix of logarithms from the simple linear factors and a reciprocal term from $x^2$.

🔍 Related terms

partial fractions, rational functions, logarithmic integration