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# 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 gaussian elimination

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

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

Simplifying

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

Expand the polynomial

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

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

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 function times a constant ($-1$) is equal to the constant times the integral of the function

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

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C$, where $b=1$ and $n=1$

$-\ln\left(x+1\right)$
14

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

$-\ln\left(x+1\right)$
15

Gather the results of all integrals

$\ln\left(x\right)-\ln\left(x+1\right)$
16

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

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

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