Step-by-step Solution

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

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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 solution

Problem to solve:

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

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

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

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

$1$
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

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

Applying the power of a power property

$\int\frac{1}{\left(x+1\right)\left(x-1\right)}dx$
6

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

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

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

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

$u=x+1$

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

$du=dx$

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

$\int\frac{-\frac{1}{2}}{u}du$

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: $x+1$

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

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

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

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

$u=x+1$

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

$du=dx$

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

$\int\frac{-\frac{1}{2}}{u}du$

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: $x+1$

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

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

$u=x-1$

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

$du=dx$

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

$\int\frac{\frac{1}{2}}{u}du$

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: $x-1$

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

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

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

Any expression multiplied by $1$ is equal to itself

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

The integral of the sum of two or more functions is equal to the sum of their integrals

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

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

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

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

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

We can solve the integral $\int\ln\left(x+1\right)dx$ 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

$u=x+1$

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

$du=dx$

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

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

The integral of the natural logarithm is given by the following formula, $\displaystyle\int\ln(x)dx=x\ln(x)-x$

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

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

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

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

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

We can solve the integral $\int\ln\left(x+1\right)dx$ 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

$u=x+1$

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

$du=dx$

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

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

The integral of the natural logarithm is given by the following formula, $\displaystyle\int\ln(x)dx=x\ln(x)-x$

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

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

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

We can solve the integral $\int\ln\left(x-1\right)dx$ 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

$u=x-1$

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

$du=dx$

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

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

The integral of the natural logarithm is given by the following formula, $\displaystyle\int\ln(x)dx=x\ln(x)-x$

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

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

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

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

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

Solve the product $-(x+1)$

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

Solve the product $-(x-1)$

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

Simplifying

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

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

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

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

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

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

Multiply $\frac{1}{2}$ times $1$

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

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

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

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

Multiply $\frac{1}{2}$ times $1$

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

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

Multiply $\frac{1}{2}$ times $-1$

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

Multiply $\frac{1}{2}$ times $-1$

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

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

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

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

Multiply $\frac{1}{2}$ times $1$

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

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

Multiply $\frac{1}{2}$ times $-1$

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

Multiply $\frac{1}{2}$ times $-1$

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

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

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

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

Multiply $\frac{1}{2}$ times $1$

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

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

Multiply $\frac{1}{2}$ times $-1$

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

Multiply $\frac{1}{2}$ times $-1$

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

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

Multiply $-\frac{1}{2}$ times $-1$

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

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

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

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

Multiply $\frac{1}{2}$ times $1$

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

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

Multiply $\frac{1}{2}$ times $-1$

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

Multiply $\frac{1}{2}$ times $-1$

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

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

Multiply $-\frac{1}{2}$ times $-1$

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

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

Multiply $-\frac{1}{2}$ times $-1$

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

Any expression multiplied by $1$ is equal to itself

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

Cancel like terms $-\frac{1}{2}x$ 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

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