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Find the integral $\int\frac{1}{\left(x^2-9\right)^2}dx$

Step-by-step Solution

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Final Answer

$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$
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Step-by-step Solution

Specify the solving method

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 $\frac{1}{2}$

$\int\frac{1}{\left(\left(x+\sqrt{9}\right)\left(\sqrt{x^2}-\sqrt{9}\right)\right)^2}dx$

The power of a product is equal to the product of it's factors raised to the same power

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

Rewrite the expression $\frac{1}{\left(x^2-9\right)^2}$ inside the integral in factored form

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

Rewrite the fraction $\frac{1}{\left(x+3\right)^2\left(x-3\right)^2}$ in $4$ simpler fractions using partial fraction decomposition

$\frac{1}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{A}{\left(x+3\right)^2}+\frac{B}{\left(x-3\right)^2}+\frac{C}{x+3}+\frac{D}{x-3}$
3

Find the values for the unknown coefficients: $A, B, C, D$. The first step is to multiply both sides of the equation from the previous step by $\left(x+3\right)^2\left(x-3\right)^2$

$1=\left(x+3\right)^2\left(x-3\right)^2\left(\frac{A}{\left(x+3\right)^2}+\frac{B}{\left(x-3\right)^2}+\frac{C}{x+3}+\frac{D}{x-3}\right)$
4

Multiplying polynomials

$1=\frac{\left(x+3\right)^2\left(x-3\right)^2A}{\left(x+3\right)^2}+\frac{\left(x+3\right)^2\left(x-3\right)^2B}{\left(x-3\right)^2}+\frac{\left(x+3\right)^2\left(x-3\right)^2C}{x+3}+\frac{\left(x+3\right)^2\left(x-3\right)^2D}{x-3}$
5

Simplifying

$1=\left(x-3\right)^2A+\left(x+3\right)^2B+\left(x+3\right)\left(x-3\right)^2C+\left(x+3\right)^2\left(x-3\right)D$
6

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

$\begin{matrix}1=36A&\:\:\:\:\:\:\:(x=-3) \\ 1=36B&\:\:\:\:\:\:\:(x=3) \\ 1=A+49B+7C+49D&\:\:\:\:\:\:\:(x=4) \\ 1=49A+B-49C-7D&\:\:\:\:\:\:\:(x=-4)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix}36A & + & 0B & + & 0C & + & 0D & =1 \\ 0A & + & 36B & + & 0C & + & 0D & =1 \\ 1A & + & 49B & + & 7C & + & 49D & =1 \\ 49A & + & 1B & - & 49C & + & 0D & =1\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}36 & 0 & 0 & 0 & 1 \\ 0 & 36 & 0 & 0 & 1 \\ 1 & 49 & 7 & 49 & 1 \\ 49 & 1 & -49 & 0 & 1\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{36} \\ 0 & 1 & 0 & 0 & \frac{1}{36} \\ 0 & 0 & 1 & 0 & \frac{1}{126} \\ 0 & 0 & 0 & 1 & -\frac{4}{441}\end{matrix}\right)$
10

The integral of $\frac{1}{\left(x+3\right)^2\left(x-3\right)^2}$ in decomposed fraction equals

$\int\left(\frac{1}{36\left(x+3\right)^2}+\frac{1}{36\left(x-3\right)^2}+\frac{\frac{1}{126}}{x+3}+\frac{-\frac{4}{441}}{x-3}\right)dx$
11

Expand the integral $\int\left(\frac{1}{36\left(x+3\right)^2}+\frac{1}{36\left(x-3\right)^2}+\frac{\frac{1}{126}}{x+3}+\frac{-\frac{4}{441}}{x-3}\right)dx$ into $4$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{36\left(x+3\right)^2}dx+\int\frac{1}{36\left(x-3\right)^2}dx+\int\frac{\frac{1}{126}}{x+3}dx+\int\frac{-\frac{4}{441}}{x-3}dx$

Take the constant $\frac{1}{36}$ out of the integral

$\frac{1}{36}\int\frac{1}{\left(x+3\right)^2}dx$

Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=3$, $c=2$ and $n=1$

$\frac{-1}{\left(2-1\right)\cdot 36\left(x+3\right)^{\left(2-1\right)}}$

Simplify the expression inside the integral

$\frac{-1}{36\left(x+3\right)}$
12

The integral $\int\frac{1}{36\left(x+3\right)^2}dx$ results in: $\frac{-1}{36\left(x+3\right)}$

$\frac{-1}{36\left(x+3\right)}$

Take the constant $\frac{1}{36}$ out of the integral

$\frac{1}{36}\int\frac{1}{\left(x-3\right)^2}dx$

Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=-3$, $c=2$ and $n=1$

$\frac{-1}{\left(2-1\right)\cdot 36\left(x-3\right)^{\left(2-1\right)}}$

Simplify the expression inside the integral

$\frac{-1}{36\left(x-3\right)}$
13

The integral $\int\frac{1}{36\left(x-3\right)^2}dx$ results in: $\frac{-1}{36\left(x-3\right)}$

$\frac{-1}{36\left(x-3\right)}$

The integral of a function times a constant ($\frac{1}{126}$) is equal to the constant times the integral of the function

$\frac{1}{126}\int\frac{1}{3+x}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=3$ and $n=1$

$\frac{1}{126}\ln\left(x+3\right)$
14

The integral $\int\frac{\frac{1}{126}}{x+3}dx$ results in: $\frac{1}{126}\ln\left(x+3\right)$

$\frac{1}{126}\ln\left(x+3\right)$

The integral of a function times a constant ($-\frac{4}{441}$) is equal to the constant times the integral of the function

$-\frac{4}{441}\int\frac{1}{-3+x}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-3$ and $n=1$

$-\frac{4}{441}\ln\left(x-3\right)$
15

The integral $\int\frac{-\frac{4}{441}}{x-3}dx$ results in: $-\frac{4}{441}\ln\left(x-3\right)$

$-\frac{4}{441}\ln\left(x-3\right)$
16

Gather the results of all integrals

$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)$
17

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

$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$

Final Answer

$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$

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Function Plot

Plotting: $\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$

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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

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.

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