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
The derivative of the linear function is equal to $1$
$1$
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
Simplify $\sqrt{x^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$
$\int\frac{1}{\left(x+\sqrt{1}\right)\left(\sqrt{x^2}-\sqrt{1}\right)}dx$
Calculate the power $\sqrt{1}$
$\int\frac{1}{\left(x+1\right)\left(\sqrt{x^2}-\sqrt{1}\right)}dx$
Simplify $\sqrt{x^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$
$\int\frac{1}{\left(x+1\right)\left(x-\sqrt{1}\right)}dx$
Calculate the power $\sqrt{1}$
$\int\frac{1}{\left(x+1\right)\left(x- 1\right)}dx$
Any expression multiplied by $1$ is equal to itself
$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
6
Factor the difference of squares $x^2-1$ as the product of two conjugated binomials
$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
Explain this step further
Intermediate steps
Rewrite the fraction $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{1}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$
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 $\left(x+1\right)\left(x-1\right)$
$1=\left(x+1\right)\left(x-1\right)\left(\frac{A}{x+1}+\frac{B}{x-1}\right)$
$1=\frac{\left(x+1\right)\left(x-1\right)A}{x+1}+\frac{\left(x+1\right)\left(x-1\right)B}{x-1}$
$1=\left(x-1\right)A+\left(x+1\right)B$
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
Proceed to solve the system of linear equations
$\begin{matrix} -2A & + & 0B & =1 \\ 0A & + & 2B & =1\end{matrix}$
Rewrite as a coefficient matrix
$\left(\begin{matrix}-2 & 0 & 1 \\ 0 & 2 & 1\end{matrix}\right)$
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
The integral of $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in decomposed fractions equals
$\frac{-1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}$
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
Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{-1}{x+1}dx+\int\frac{1}{2\left(x-1\right)}dx$
9
Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{-1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$
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
Expand the integral $\int\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
$\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)x- \left(-\frac{1}{2}\right)\int\ln\left(x+1\right)dx- \left(\frac{1}{2}\right)\int\ln\left(x-1\right)dx$
Multiply the fraction and term in $- \left(-\frac{1}{2}\right)\int\ln\left(x+1\right)dx$
$\left(-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|\right)x+\frac{1}{2}\int\ln\left(x+1\right)dx- \left(\frac{1}{2}\right)\int\ln\left(x-1\right)dx$
Multiply the fraction and term in $- \left(\frac{1}{2}\right)\int\ln\left(x-1\right)dx$
$\left(-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|\right)x+\frac{1}{2}\int\ln\left(x+1\right)dx-\frac{1}{2}\int\ln\left(x-1\right)dx$
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
The integral $\int\ln\left(x+1\right)dx$ results in $\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
15
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
Simplify the product $-(x+1)$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-\left(x-1\right)\right)$
Simplify the product $-(x-1)$
$-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)$
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 $-\frac{1}{2}x\ln\left|x+1\right|+\frac{1}{2}x\ln\left|x-1\right|+\frac{1}{2}\left(\left(x+1\right)\ln\left|x+1\right|-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left|x-1\right|-x+1\right)+C_0$
