Step-by-step Solution

Integral of $\frac{x}{x^2-1}$ with respect to x

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Final Answer

$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$

Step-by-step explanation

Problem to solve:

$\int\frac{x}{x^2-1}dx$

Choose the solving method

1

Rewrite the fraction $\frac{x}{x^2-1}$ inside the integral as a product: $x\frac{1}{x^2-1}$

$\int x\frac{1}{x^2-1}dx$
2

We can solve the integral $\int\frac{x}{x^2-1}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
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

$v=\int\frac{1}{x^2-1}dx$
6

Factor the difference of squares $x^2-1$ as the product of two conjugated binomials

$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
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

$1=\frac{A\left(x+1\right)\left(x-1\right)}{x+1}+\frac{B\left(x+1\right)\left(x-1\right)}{x-1}$
10

Simplifying

$1=A\left(x-1\right)+B\left(x+1\right)$
11

Expand the polynomial

$1=Ax-A+Bx+B$
12

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
13

Proceed to solve the system of linear equations

$\begin{matrix} -2A & + & 0B & =1 \\ 0A & + & 2B & =1\end{matrix}$
14

Rewrite as a coefficient matrix

$\left(\begin{matrix}-2 & 0 & 1 \\ 0 & 2 & 1\end{matrix}\right)$
15

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
16

The integral of $\frac{1}{\left(x+1\right)\left(x-1\right)}$ in decomposed fraction equals

$\int\left(\frac{-\frac{1}{2}}{x+1}+\frac{\frac{1}{2}}{x-1}\right)dx$
17

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=1$ and $n=-\frac{1}{2}$

$-\frac{1}{2}\ln\left|x+1\right|+\int\frac{\frac{1}{2}}{x-1}dx$
18

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-1$ and $n=\frac{1}{2}$

$-\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|$
19

Now replace the values of $u$, $du$ and $v$ in the last formula

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)-\left(\int-\frac{1}{2}\ln\left(x+1\right)dx+\int\frac{1}{2}\ln\left(x-1\right)dx\right)$
20

Multiplying polynomials $-1$ and $-\frac{1}{2}\int\ln\left(x+1\right)dx+\frac{1}{2}\int\ln\left(x-1\right)dx$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\int\ln\left(x+1\right)dx-\frac{1}{2}\int\ln\left(x-1\right)dx$
21

The integral $\int\ln\left(x+1\right)dx$ results in $\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\int\ln\left(x-1\right)dx$
22

The integral $\int\ln\left(x-1\right)dx$ results in $\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)$
23

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)+C_0$
24

Solve the product $\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(x+1\right)\ln\left(x+1\right)-\frac{1}{2}\left(x+1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)+C_0$
25

Solve the product $\frac{1}{2}\left(x+1\right)\ln\left(x+1\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}\left(x+1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)+C_0$
26

Solve the product $-\frac{1}{2}\left(x+1\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)+C_0$
27

Solve the product $-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-\left(x-1\right)\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}-\frac{1}{2}\left(x-1\right)\ln\left(x-1\right)+\frac{1}{2}\left(x-1\right)+C_0$
28

Solve the product $-\frac{1}{2}\left(x-1\right)\ln\left(x-1\right)$

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x-\frac{1}{2}+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+\frac{1}{2}\left(x-1\right)+C_0$
29

Solve the product $\frac{1}{2}\left(x-1\right)$

$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)-\frac{1}{2}x+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+\frac{1}{2}x+C_0$
30

Adding $-\frac{1}{2}x$ and $\frac{1}{2}x$

$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$

Final Answer

$-1-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\left(\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x+1\right)+\left(-\frac{1}{2}x+\frac{1}{2}\right)\ln\left(x-1\right)+C_0$