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

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##  Final answer to the problem

$\frac{1}{2}\ln\left|x+1\right|+\frac{1}{2}\ln\left|x-1\right|+C_0$
Got another answer? Verify it here!

##  Step-by-step Solution 

How should I solve this problem?

• Integrate by partial fractions
• Integrate by substitution
• Integrate by parts
• Integrate using tabular integration
• Integrate by trigonometric substitution
• Weierstrass Substitution
• Integrate using trigonometric identities
• Integrate using basic integrals
• Product of Binomials with Common Term
• FOIL Method
Can't find a method? Tell us so we can add it.
1

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

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

$\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$
3

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

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

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

$x=\left(x-1\right)A+\left(x+1\right)B$
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Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}-1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
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Proceed to solve the system of linear equations

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

$\left(\begin{matrix}-2 & 0 & -1 \\ 0 & 2 & 1\end{matrix}\right)$
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Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
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The integral of $\frac{x}{\left(x+1\right)\left(x-1\right)}$ in decomposed fraction equals

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

$\int\frac{1}{2\left(x+1\right)}dx+\int\frac{1}{2\left(x-1\right)}dx$
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We can solve the integral $\int\frac{1}{2\left(x+1\right)}dx$ 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

$u=x+1$
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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

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

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

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

$\frac{1}{2}\ln\left|x-1\right|$
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Gather the results of all integrals

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

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

##  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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7
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9
0
a
b
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

###  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.