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

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$-\frac{1}{2}x\ln\left(x+1\right)+\frac{1}{2}x\ln\left(x-1\right)+\frac{1}{2}\left(\left(x+1\right)\ln\left(x+1\right)-x-1\right)-\frac{1}{2}\left(\left(x-1\right)\ln\left(x-1\right)-x+1\right)+C_0$
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 Step-by-step Solution 

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

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

Simplify $\sqrt{x^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$

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

Calculate the power $\sqrt{1}$

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

Simplify $\sqrt{x^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$

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

Calculate the power $\sqrt{1}$

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

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

Simplifying

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

Expand the polynomial

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

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

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int\frac{-1}{x+1}dx+\int\frac{1}{2\left(x-1\right)}dx$
18

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int\frac{-1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$
19

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C, where b=1 and n=-1 -\frac{1}{2}\ln\left(x+1\right)+\frac{1}{2}\int\frac{1}{x-1}dx 20 Apply the formula: \int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$

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

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

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

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

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

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

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

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

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

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

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

Simplify the product $-(x+1)$

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

Simplify the product $-(x-1)$

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

Simplify the expression inside the integral

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

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

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

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

 Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

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u
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x
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.
(◻)
+
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e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Main Topic: Integrals of Rational Functions

Integrals of rational functions of the form R(x) = P(x)/Q(x).

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