Try NerdPal! Our new app on iOS and Android

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

## Step-by-step Solution

Go!
Go!
1
2
3
4
5
6
7
8
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

### Videos

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

## Step-by-step Solution

Problem to solve:

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

Specify the solving method

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

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

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

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

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

$x=\left(x+1\right)\left(x-1\right)\left(\frac{A}{x+1}+\frac{B}{x-1}\right)$
4

Multiplying polynomials

$x=\frac{A\left(x+1\right)\left(x-1\right)}{x+1}+\frac{B\left(x+1\right)\left(x-1\right)}{x-1}$
5

Simplifying

$x=A\left(x-1\right)+B\left(x+1\right)$
6

Expand the polynomial

$x=Ax-A+Bx+B$
7

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

$\begin{matrix}-1=-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=2B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix} -2A & + & 0B & =-1 \\ 0A & + & 2B & =1\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}-2 & 0 & -1 \\ 0 & 2 & 1\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2}\end{matrix}\right)$
11

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

$\int\left(\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}\right)dx$
12

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$$=n\ln\left(x+b\right)+C, where b=1 and n=1 \frac{1}{2}\ln\left(x+1\right) 13 The integral \int\frac{1}{2\left(x+1\right)}dx results in: \frac{1}{2}\ln\left(x+1\right) \frac{1}{2}\ln\left(x+1\right) Take the constant \frac{1}{2} out of the integral \frac{1}{2}\int\frac{1}{x-1}dx Apply the formula: \int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$

$\frac{1}{2}\ln\left(x-1\right)$
14

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

$\frac{1}{2}\ln\left(x-1\right)$
15

Gather the results of all integrals

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

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

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

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

### beta Got another answer? Verify it!

Go!
1
2
3
4
5
6
7
8
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

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

### Main topic:

Integrals by Partial Fraction expansion

~ 0.12 s