Final answer to the problem
$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$
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Step-by-step Solution
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Intermediate steps
1
Rewrite the expression $\frac{1}{\left(x^2-9\right)^2}$ inside the integral in factored form
$\int\frac{1}{\left(x+3\right)^2\left(x-3\right)^2}dx$
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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
$1=\frac{\left(x+3\right)^2\left(x-3\right)^2A}{\left(x+3\right)^2}+\frac{\left(x+3\right)^2\left(x-3\right)^2B}{\left(x-3\right)^2}+\frac{\left(x+3\right)^2\left(x-3\right)^2C}{x+3}+\frac{\left(x+3\right)^2\left(x-3\right)^2D}{x-3}$
$1=\left(x-3\right)^2A+\left(x+3\right)^2B+\left(x+3\right)\left(x-3\right)^2C+\left(x+3\right)^2\left(x-3\right)D$
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
$\int\frac{1}{36\left(x+3\right)^2}dx+\int\frac{1}{36\left(x-3\right)^2}dx+\int\frac{\frac{1}{126}}{x+3}dx+\int\frac{-\frac{4}{441}}{x-3}dx$
Intermediate steps
12
The integral $\int\frac{1}{36\left(x+3\right)^2}dx$ results in: $\frac{-1}{36\left(x+3\right)}$
$\frac{-1}{36\left(x+3\right)}$
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Intermediate steps
13
The integral $\int\frac{1}{36\left(x-3\right)^2}dx$ results in: $\frac{-1}{36\left(x-3\right)}$
$\frac{-1}{36\left(x-3\right)}$
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Intermediate steps
14
The integral $\int\frac{\frac{1}{126}}{x+3}dx$ results in: $\frac{1}{126}\ln\left(x+3\right)$
$\frac{1}{126}\ln\left(x+3\right)$
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Intermediate steps
15
The integral $\int\frac{-\frac{4}{441}}{x-3}dx$ results in: $-\frac{4}{441}\ln\left(x-3\right)$
$-\frac{4}{441}\ln\left(x-3\right)$
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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 to the problem
$\frac{-1}{36\left(x+3\right)}+\frac{-1}{36\left(x-3\right)}+\frac{1}{126}\ln\left(x+3\right)-\frac{4}{441}\ln\left(x-3\right)+C_0$