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

Step-by-step Solution

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Final answer to the problem

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

How should I solve this problem?

  • Integrate by parts
  • Integrate by partial fractions
  • Integrate by substitution
  • 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
  • Load more...
Can't find a method? Tell us so we can add it.
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

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
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 to find $v$

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

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

$\begin{matrix}1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
12

Proceed to solve the system of linear equations

$\begin{matrix} -2A & + & 0B & =1 \\ 0A & + & 2B & =1\end{matrix}$
13

Rewrite as a coefficient matrix

$\left(\begin{matrix}-2 & 0 & 1 \\ 0 & 2 & 1\end{matrix}\right)$
14

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
15

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

$\int\left(\frac{-1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}\right)dx$
16

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

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

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

$\frac{1}{2}\int\frac{-1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$
18

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the expression inside the integral

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

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

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

Final answer to the problem

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

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Function Plot

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

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a
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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

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