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Integrals by partial fraction expansion Calculator

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1

Solved example of integrals by partial fraction expansion

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

Rewrite the fraction $\frac{1}{x\left(x+1\right)}$ in $2$ simpler fractions using partial fraction decomposition

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

Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $x\left(x+1\right)$

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

Multiplying polynomials

$1=\frac{xA\left(x+1\right)}{x}+\frac{xB\left(x+1\right)}{x+1}$
5

Simplifying

$1=A\left(x+1\right)+xB$
6

Expand the polynomial

$1=Ax+A+xB$
7

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

$\begin{matrix}1=A&\:\:\:\:\:\:\:(x=0) \\ 1=-B&\:\:\:\:\:\:\:(x=-1)\end{matrix}$
8

Proceed to solve the system of linear equations

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

Rewrite as a coefficient matrix

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

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

The integral of $\frac{1}{x\left(x+1\right)}$ in decomposed fraction equals

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

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

$\int\frac{1}{x}dx+\int\frac{-1}{x+1}dx$

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

$\ln\left|x\right|$
13

The integral $\int\frac{1}{x}dx$ results in: $\ln\left|x\right|$

$\ln\left|x\right|$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$-\int\frac{1}{1+x}dx$

Solve the integral $\int\frac{1}{1+x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=1+x \\ du=dx\end{matrix}$

Substituting $u$ and $dx$ in the integral and simplify

$-\int\frac{1}{u}du$

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

$-\ln\left|u\right|$

Substitute $u$ back for it's value, $1+x$

$-\ln\left|1+x\right|$
14

The integral $\int\frac{-1}{x+1}dx$ results in: $-\ln\left|1+x\right|$

$-\ln\left|1+x\right|$
15

After gathering the results of all integrals, the final answer is

$\ln\left|x\right|-\ln\left|1+x\right|$
16

As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration

$\ln\left|x\right|-\ln\left|1+x\right|+C_0$

Answer

$\ln\left|x\right|-\ln\left|1+x\right|+C_0$

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