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

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Final answer to the problem

$\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
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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 $\frac{1}{2}$

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

Calculate the power $\sqrt{1}$

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

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 $\frac{1}{2}$

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

Calculate the power $\sqrt{1}$

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

Any expression multiplied by $1$ is equal to itself

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

Rewrite the expression $\frac{x}{x^2-1}$ inside the integral in factored form

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

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

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

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

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

Multiplying polynomials

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

Simplifying

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

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

$\begin{matrix}-1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$

Proceed to solve the system of linear equations

$\begin{matrix} -2A & + & 0B & =-1 \\ 0A & + & 2B & =1\end{matrix}$

Rewrite as a coefficient matrix

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

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

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

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

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

$\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}$
3

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

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

$\frac{1}{2}\ln\left|x+1\right|$
4

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

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

$\frac{1}{2}\ln\left|x-1\right|$
5

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

$\frac{1}{2}\ln\left|x-1\right|$
6

Gather the results of all integrals

$\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|$
7

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 to the problem

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

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Function Plot

Plotting: $\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|+C_0$

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a
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c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

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

How to improve your answer:

Main Topic: Integrals by Partial Fraction Expansion

The partial fraction decomposition or partial fraction expansion of a rational function is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

Used Formulas

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