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
Intermediate steps
7
Rewrite the fraction $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{-1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}$
Explain this step further
8
Apply the formula: $\int\frac{a}{bc}dx$$=\frac{1}{c}\int\frac{a}{b}dx$, where $a=-1$, $b=x+1$ and $c=2$
$\frac{1}{2}\int\frac{-1}{x+1}dx+\int\frac{1}{2\left(x-1\right)}dx$
9
Apply the formula: $\int\frac{a}{bc}dx$$=\frac{1}{c}\int\frac{a}{b}dx$, where $a=1$, $b=x-1$ and $c=2$
$\frac{1}{2}\int\frac{-1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$
10
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$ and $n=-1$
$-\left(\frac{1}{2}\right)\ln\left(x+1\right)+\frac{1}{2}\int\frac{1}{x-1}dx$
11
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$
$-\frac{1}{2}\ln\left(x+1\right)+1\left(\frac{1}{2}\right)\ln\left(x-1\right)$
Intermediate steps
12
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
13
Multiply the single term $x$ by each term of the polynomial $\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)$
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\int\ln\left(x+1\right)dx-\frac{1}{2}\int\ln\left(x-1\right)dx$
14
Apply the formula: $\int\ln\left(x+b\right)dx$$=\left(x+b\right)\ln\left(x+b\right)-\left(x+b\right)+C$, where $b=1$ and $x+b=x+1$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
15
Apply the formula: $\int\ln\left(x+b\right)dx$$=\left(x+b\right)\ln\left(x+b\right)-\left(x+b\right)+C$, where $b=-1$ and $x+b=x-1$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-\left(x-1\right)\right)$
Intermediate steps
16
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
17
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 $-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)+C_0$
