# Integrals by partial fraction expansion Calculator

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### Difficult Problems

1

Example

$\int\frac{1}{\left(x+2\right)\left(x-3\right)}dx$
2

Using partial fraction decomposition, the fraction $\frac{1}{\left(x-3\right)\left(2+x\right)}$ can be rewritten as

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

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x-3\right)\left(2+x\right)$

$1=\left(\frac{A}{x-3}+\frac{B}{2+x}\right)\left(x-3\right)\left(2+x\right)$
4

Multiplying polynomials

$1=\frac{A\left(x-3\right)\left(2+x\right)}{x-3}+\frac{B\left(x-3\right)\left(2+x\right)}{2+x}$
5

Simplifying

$1=A\left(2+x\right)+B\left(x-3\right)$
6

Expand the polynomial

$1=2A+Ax-3B+Bx$
7

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

$\begin{matrix}1=-5B&\:\:\:\:\:\:\:(x=-2) \\ 1=-6B-A&\:\:\:\:\:\:\:(x=-3)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix}0A & - & 5B & =1 \\ -1A & - & 6B & =1\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & -5 & 1 \\ -1 & -6 & 1\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & \frac{1}{5} \\ 0 & 1 & -\frac{1}{5}\end{matrix}\right)$
11

The decomposed integral equivalent is

$\int\left(\frac{\frac{1}{5}}{x-3}+\frac{-\frac{1}{5}}{2+x}\right)dx$
12

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

$\int\frac{\frac{1}{5}}{x-3}dx+\int\frac{-\frac{1}{5}}{2+x}dx$
13

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|, where b=-3 and n=\frac{1}{5} \frac{1}{5}\ln\left|x-3\right|+\int\frac{-\frac{1}{5}}{2+x}dx 14 Apply the formula: \int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=2$ and $n=-\frac{1}{5}$

$\frac{1}{5}\ln\left|x-3\right|-\frac{1}{5}\ln\left|2+x\right|$
15

$\frac{1}{5}\ln\left|x-3\right|-\frac{1}{5}\ln\left|2+x\right|+C_0$