# Integral of 1/(x^2(x+2))

## \int\frac{1}{x^2\left(x+2\right)}dx

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$\frac{1}{4}\ln\left|2+x\right|+\frac{-\frac{1}{2}}{x}-\frac{1}{4}\ln\left|x\right|+C_0$

## Step by step solution

Problem

$\int\frac{1}{x^2\left(x+2\right)}dx$
1

Using partial fraction decomposition, the fraction $\frac{1}{\left(2+x\right)x^2}$ can be rewritten as

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

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(2+x\right)x^2$

$1=\left(\frac{A}{2+x}+\frac{B}{x^2}+\frac{C}{x}\right)\left(2+x\right)x^2$
3

Multiplying polynomials

$1=\frac{A\left(2+x\right)x^2}{2+x}+\frac{B\left(2+x\right)x^2}{x^2}+\frac{C\left(2+x\right)x^2}{x}$
4

Simplifying

$1=Ax^2+B\left(2+x\right)+Cx\left(2+x\right)$
5

Expand the polynomial

$1=Ax^2+2B+Bx+2Cx+Cx^2$
6

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}1=2B&\:\:\:\:\:\:\:(x=0) \\ 1=4A+2B-2B-4C+4C&\:\:\:\:\:\:\:(x=-2) \\ 1=4A+4B+8C&\:\:\:\:\:\:\:(x=2)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix}0A & + & 2B & + & 0C & =1 \\ 4A & + & 0B & + & 0C & =1 \\ 4A & + & 4B & + & 8C & =1\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & 2 & 0 & 1 \\ 4 & 0 & 0 & 1 \\ 4 & 4 & 8 & 1\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & \frac{1}{4} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & -\frac{1}{4}\end{matrix}\right)$
10

The decomposed integral equivalent is

$\int\left(\frac{\frac{1}{4}}{2+x}+\frac{\frac{1}{2}}{x^2}+\frac{-\frac{1}{4}}{x}\right)dx$
11

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int\frac{\frac{1}{4}}{2+x}dx+\int\frac{\frac{1}{2}}{x^2}dx+\int\frac{-\frac{1}{4}}{x}dx$
12

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\int\frac{\frac{1}{4}}{2+x}dx+\int\frac{\frac{1}{2}}{x^2}dx-\frac{1}{4}\ln\left|x\right|$
13

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|, where b=2 and n=\frac{1}{4} \frac{1}{4}\ln\left|2+x\right|+\int\frac{\frac{1}{2}}{x^2}dx-\frac{1}{4}\ln\left|x\right| 14 Rewrite the exponent using the power rule \frac{a^m}{a^n}=a^{m-n}, where in this case m=0 \frac{1}{4}\ln\left|2+x\right|+\int\frac{1}{2}x^{-2}dx-\frac{1}{4}\ln\left|x\right| 15 Taking the constant out of the integral \frac{1}{4}\ln\left|2+x\right|+\frac{1}{2}\int x^{-2}dx-\frac{1}{4}\ln\left|x\right| 16 Apply the power rule for integration, \displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}, where n represents a constant function \frac{1}{4}\ln\left|2+x\right|+\frac{1}{2}\cdot\frac{x^{-1}}{-1}-\frac{1}{4}\ln\left|x\right| 17 Simplify the fraction \frac{1}{4}\ln\left|2+x\right|-\frac{1}{2}x^{-1}-\frac{1}{4}\ln\left|x\right| 18 Applying the property of exponents, \displaystyle a^{-n}=\frac{1}{a^n}, where n is a number \frac{1}{4}\ln\left|2+x\right|-\frac{1}{2}\cdot\frac{1}{x}-\frac{1}{4}\ln\left|x\right| 19 Apply the formula: a\frac{1}{x}$$=\frac{a}{x}$, where $a=-\frac{1}{2}$

$\frac{1}{4}\ln\left|2+x\right|+\frac{-\frac{1}{2}}{x}-\frac{1}{4}\ln\left|x\right|$
20

$\frac{1}{4}\ln\left|2+x\right|+\frac{-\frac{1}{2}}{x}-\frac{1}{4}\ln\left|x\right|+C_0$

$\frac{1}{4}\ln\left|2+x\right|+\frac{-\frac{1}{2}}{x}-\frac{1}{4}\ln\left|x\right|+C_0$

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Integrals by partial fraction expansion

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