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
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Integrate by parts Integrate by partial fractions Integrate by substitution Integrate using tabular integration Integrate by trigonometric substitution Weierstrass Substitution Integrate using trigonometric identities Integrate using basic integrals Product of Binomials with Common Term FOIL Method
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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 $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
Explain this step further
4
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=\frac{1}{x^2-1}dx}\\ \displaystyle{\int dv=\int \frac{1}{x^2-1}dx}\end{matrix}$
$v=\int\frac{1}{x^2-1}dx$
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$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$ and $n=-1$
$-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\int\frac{1}{x-1}dx$
19
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$
$-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)$
Intermediate steps
20
Now replace the values of $u$, $du$ and $v$ in the last formula
$\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)x-\int-\frac{1}{2}\ln\left(x+1\right)dx-\int\frac{1}{2}\ln\left(x-1\right)dx$
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
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
23
The integral $\int\ln\left(x-1\right)dx$ results in
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)$
Intermediate steps
24
Simplify the expression inside the integral
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)$
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
$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$