Final answer to the problem
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)+C_0$
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Step-by-step Solution
1
Rewrite the fraction $\frac{x}{x^2-1}$ inside the integral as the product of two functions: $x\frac{1}{x^2-1}$
$\int x\frac{1}{x^2-1}dx$
2
We can solve the integral $\int x\frac{1}{x^2-1}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
3
First, identify or choose $u$ and calculate it's derivative, $du$
$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
Explain this step further
4
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=\frac{1}{x^2-1}dx}\\ \displaystyle{\int dv=\int \frac{1}{x^2-1}dx}\end{matrix}$
5
Solve the integral to find $v$
$v=\int\frac{1}{x^2-1}dx$
Intermediate steps
6
Factor the difference of squares $x^2-1$ as the product of two conjugated binomials
$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
Explain this step further
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
$1=\frac{\left(x+1\right)\left(x-1\right)A}{x+1}+\frac{\left(x+1\right)\left(x-1\right)B}{x-1}$
$1=\left(x-1\right)A+\left(x+1\right)B$
11
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
12
Proceed to solve the system of linear equations
$\begin{matrix} -2A & + & 0B & =1 \\ 0A & + & 2B & =1\end{matrix}$
13
Rewrite as a coefficient matrix
$\left(\begin{matrix}-2 & 0 & 1 \\ 0 & 2 & 1\end{matrix}\right)$
14
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
15
The integral of $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in decomposed fraction equals
$\int\left(\frac{-1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}\right)dx$
16
Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{-1}{x+1}dx+\int\frac{1}{2\left(x-1\right)}dx$
17
Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{-1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$
18
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$ and $n=-1$
$\left(\frac{1}{2}\right)\left(-1\right)\ln\left(x+1\right)+\frac{1}{2}\int\frac{1}{x-1}dx$
19
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$
$-\frac{1}{2}\ln\left(x+1\right)+\left(\frac{1}{2}\right)\cdot 1\ln\left(x-1\right)$
Intermediate steps
20
Now replace the values of $u$, $du$ and $v$ in the last formula
$\left(-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|\right)x+\frac{1}{2}\int\ln\left(x+1\right)dx-\frac{1}{2}\int\ln\left(x-1\right)dx$
Explain this step further
21
Multiply the single term $x$ by each term of the polynomial $\left(-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\int\ln\left(x+1\right)dx-\frac{1}{2}\int\ln\left(x-1\right)dx$
22
The integral $\int\ln\left(x+1\right)dx$ results in $\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
23
The integral $\int\ln\left(x-1\right)dx$ results in $\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-\left(x-1\right)\right)$
Intermediate steps
24
Simplify the expression
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)$
Explain this step further
25
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)+C_0$
Final answer to the problem
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)+C_0$