# Integrals by Partial Fraction expansion Calculator

## Get detailed solutions to your math problems with our Integrals by Partial Fraction expansion step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here!

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### Difficult Problems

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 for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step 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

Expand the integral $\int\left(\frac{1}{x}+\frac{-1}{x+1}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

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

We can solve the integral $\int\frac{-1}{x+1}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x+1$

Differentiate both sides of the equation $u=x+1$

$du=\frac{d}{dx}\left(x+1\right)$

Find the derivative

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

The derivative of a sum of two or more functions is the sum of the derivatives of each function

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

The derivative of the constant function ($1$) is equal to zero

$\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$1$
14

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=dx$
15

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

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

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

$\ln\left(x\right)$
16

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

$\ln\left(x\right)$

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

$-\ln\left(u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $x+1$

$-\ln\left(x+1\right)$
17

The integral $\int\frac{-1}{u}du$ results in: $-\ln\left(x+1\right)$

$-\ln\left(x+1\right)$
18

Gather the results of all integrals

$\ln\left(x\right)-\ln\left(x+1\right)$
19

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

$\ln\left(x\right)-\ln\left(x+1\right)+C_0$

$\ln\left(x\right)-\ln\left(x+1\right)+C_0$