$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$
Basic Integrals
Integration by substitution
Integration by parts
Integration by trigonometric substitution
Integrals by partial fraction expansion
Tabular Integration
1
Rewrite the fraction $\frac{x}{x^2-1}$ inside the integral as a product: $x\frac{1}{x^2-1}$
$\int x\frac{1}{x^2-1}dx$
2
We can solve the integral $\int\frac{x}{x^2-1}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
4
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=\frac{1}{x^2-1}dx}\\ \displaystyle{\int dv=\int \frac{1}{x^2-1}dx}\end{matrix}$
$v=\int\frac{1}{x^2-1}dx$
6
Factor the difference of squares $x^2-1$ as the product of two conjugated binomials
$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
7
Rewrite the fraction $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{1}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$
8
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 $\left(x+1\right)\left(x-1\right)$
$1=\left(x+1\right)\left(x-1\right)\left(\frac{A}{x+1}+\frac{B}{x-1}\right)$
9
Multiplying polynomials
$1=\frac{A\left(x+1\right)\left(x-1\right)}{x+1}+\frac{B\left(x+1\right)\left(x-1\right)}{x-1}$
$1=A\left(x-1\right)+B\left(x+1\right)$
12
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
13
Proceed to solve the system of linear equations
$\begin{matrix} -2A & + & 0B & =1 \\ 0A & + & 2B & =1\end{matrix}$
14
Rewrite as a coefficient matrix
$\left(\begin{matrix}-2 & 0 & 1 \\ 0 & 2 & 1\end{matrix}\right)$
15
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
16
The integral of $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in decomposed fraction equals
$\int\left(\frac{-\frac{1}{2}}{x+1}+\frac{\frac{1}{2}}{x-1}\right)dx$
17
Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=1$ and $n=-\frac{1}{2}$
$-\frac{1}{2}\ln\left|x+1\right|+\int\frac{\frac{1}{2}}{x-1}dx$
18
Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-1$ and $n=\frac{1}{2}$
$-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|$
19
Now replace the values of $u$, $du$ and $v$ in the last formula
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)-\int-\frac{1}{2}\ln\left(x+1\right)dx-\int\frac{1}{2}\ln\left(x-1\right)dx$
20
The integral $\int\ln\left(x+1\right)dx$ results in $\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
21
The integral $\int\ln\left(x-1\right)dx$ results in $\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)$
23
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
24
Solve the product $\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(x+1\right)\ln\left(x+1\right)+\frac{1}{2}\left(-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
25
Solve the product $\frac{1}{2}\left(x+1\right)\ln\left(x+1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\frac{1}{2}\left(-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
26
Solve the product $\frac{1}{2}\left(-x-1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
27
Solve the product $-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}-\frac{1}{2}\left(x-1\right)\ln\left(x-1\right)-\frac{1}{2}\left(-x+1\right)+C_0$
28
Solve the product $-\frac{1}{2}\left(x-1\right)\ln\left(x-1\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)-\frac{1}{2}\left(-x+1\right)+C_0$
29
Solve the product $-\frac{1}{2}\left(-x+1\right)$
$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+\frac{1}{2}x+C_0$
30
Cancel like terms $-\frac{1}{2}x$ and $\frac{1}{2}x$
$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$
$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$