Final Answer
$\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)+C_0$
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Step-by-step Solution
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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 $\frac{1}{2}$
$\frac{x}{\left(x+\sqrt{1}\right)\left(\sqrt{x^2}-\sqrt{1}\right)}$
Calculate the power $\sqrt{1}$
$\frac{x}{\left(x+1\right)\left(\sqrt{x^2}-\sqrt{1}\right)}$
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 $\frac{1}{2}$
$\frac{x}{\left(x+1\right)\left(x-\sqrt{1}\right)}$
Calculate the power $\sqrt{1}$
$\frac{x}{\left(x+1\right)\left(x-1\right)}$
1
Rewrite the expression $\frac{x}{x^2-1}$ inside the integral in factored form
$\int\frac{x}{\left(x+1\right)\left(x-1\right)}dx$
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2
Rewrite the fraction $\frac{x}{\left(x+1\right)\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$
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4
Multiplying polynomials
$x=\frac{\left(x+1\right)\left(x-1\right)A}{x+1}+\frac{\left(x+1\right)\left(x-1\right)B}{x-1}$
$x=\left(x-1\right)A+\left(x+1\right)B$
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7
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}-1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
8
Proceed to solve the system of linear equations
$\begin{matrix} -2A & + & 0B & =-1 \\ 0A & + & 2B & =1\end{matrix}$
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10
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
11
The integral of $\frac{x}{\left(x+1\right)\left(x-1\right)}$ in decomposed fraction equals
$\int\left(\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}\right)dx$
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Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{1}{x+1}dx$
Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C$, where $b=1$ and $n=1$
$\frac{1}{2}\ln\left(x+1\right)$
13
The integral $\int\frac{1}{2\left(x+1\right)}dx$ results in: $\frac{1}{2}\ln\left(x+1\right)$
$\frac{1}{2}\ln\left(x+1\right)$
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Take the constant $\frac{1}{2}$ out of the integral
$\frac{1}{2}\int\frac{1}{x-1}dx$
Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$
$\frac{1}{2}\ln\left(x-1\right)$
14
The integral $\int\frac{1}{2\left(x-1\right)}dx$ results in: $\frac{1}{2}\ln\left(x-1\right)$
$\frac{1}{2}\ln\left(x-1\right)$
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16
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)+C_0$
Final Answer$\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)+C_0$