Step-by-step Solution

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

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Final Answer

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

Step-by-step Solution

Problem to solve:

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

Solving method

Applying the power of a power property

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

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

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

Rewrite the fraction $\frac{x}{\left(x+1\right)\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$
3

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 $\left(x+1\right)\left(x-1\right)$

$x=\left(x+1\right)\left(x-1\right)\left(\frac{A}{x+1}+\frac{B}{x-1}\right)$
4

Multiplying polynomials

$x=\frac{A\left(x+1\right)\left(x-1\right)}{x+1}+\frac{B\left(x+1\right)\left(x-1\right)}{x-1}$
5

Simplifying

$x=A\left(x-1\right)+B\left(x+1\right)$
6

Expand the polynomial

$x=Ax-A+Bx+B$
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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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Rewrite as a coefficient matrix

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

$\int\left(\frac{\frac{1}{2}}{x+1}+\frac{\frac{1}{2}}{x-1}\right)dx$
12

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

$\int\frac{\frac{1}{2}}{x+1}dx+\int\frac{\frac{1}{2}}{x-1}dx$
13

We can solve the integral $\int\frac{\frac{1}{2}}{x+1}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x+1$

Differentiate both sides of the equation $u=x+1$

$du=\frac{d}{dx}\left(x+1\right)$

Find the derivative

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

The derivative of a sum of two functions is the sum of the derivatives of each function

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

The derivative of the constant function ($1$) is equal to zero

$\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$1$
14

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=dx$
15

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{\frac{1}{2}}{u}du+\int\frac{\frac{1}{2}}{x-1}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{1}{2}\ln\left(u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $x+1$

$\frac{1}{2}\ln\left(x+1\right)$
16

The integral $\int\frac{\frac{1}{2}}{u}du$ results in: $\frac{1}{2}\ln\left(x+1\right)$

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

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

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

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

$\frac{1}{2}\ln\left(x-1\right)$
17

The integral $\int\frac{\frac{1}{2}}{x-1}dx$ results in: $\frac{1}{2}\ln\left(x-1\right)$

$\frac{1}{2}\ln\left(x-1\right)$
18

Gather the results of all integrals

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

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

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