# Step-by-step Solution

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

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asin
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asinh
acosh
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asech
acsch

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

## Step-by-step Solution

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

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

Expand the integral $\int\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

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

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

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

Multiply the single term $x$ by each term of the polynomial $\left(-\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\ln\left(x-1\right)\right)$

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

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: $x+1$

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

Solve the product $-(x+1)$

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

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

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

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: $x+1$

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

Solve the product $-(x+1)$

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

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

$u=x-1$

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: $x-1$

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

Solve the product $-(x-1)$

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

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

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

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 $-(x-1)$

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

Simplifying

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

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

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

$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
SnapXam A2

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x
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.
(◻)
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2

e
π
ln
log
log
lim
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Dx
|◻|
θ
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<
>=
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sin
cos
tan
cot
sec
csc

asin
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atan
acot
asec
acsc

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coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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

### Main topic:

Integrals of Rational Functions

~ 0.17 s