## Final Answer

## Step-by-step Solution

Problem to solve:

Solving method

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

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

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

First, identify $u$ and calculate $du$

Now, identify $dv$ and calculate $v$

Solve the integral

Applying the power of a power property

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

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

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

Multiplying polynomials

Simplifying

Expand the polynomial

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

Proceed to solve the system of linear equations

Rewrite as a coefficient matrix

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

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

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

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

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

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

Solve the product $-(x+1)$

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

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

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

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

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

Solve the product $-(x+1)$

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

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

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

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

Solve the product $-(x-1)$

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

Solve the product $-(x+1)$

Solve the product $-(x-1)$

Simplifying

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