Final Answer
$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$
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Step-by-step Solution
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Intermediate steps
1
Rewrite the expression $\frac{x+8}{x^6-2x^4+x^2}$ inside the integral in factored form
$\int\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}dx$
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2
Rewrite the fraction $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in $6$ simpler fractions using partial fraction decomposition
$\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}=\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x^2}+\frac{D}{x-1}+\frac{F}{x+1}+\frac{G}{x}$
3
Find the values for the unknown coefficients: $A, B, C, D, F, G$. The first step is to multiply both sides of the equation from the previous step by $\left(x-1\right)^2\left(x+1\right)^2x^2$
$x+8=\left(x-1\right)^2\left(x+1\right)^2x^2\left(\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x^2}+\frac{D}{x-1}+\frac{F}{x+1}+\frac{G}{x}\right)$
4
Multiplying polynomials
$x+8=\frac{\left(x-1\right)^2\left(x+1\right)^2x^2A}{\left(x-1\right)^2}+\frac{\left(x-1\right)^2\left(x+1\right)^2x^2B}{\left(x+1\right)^2}+\frac{\left(x-1\right)^2\left(x+1\right)^2x^2C}{x^2}+\frac{\left(x-1\right)^2\left(x+1\right)^2x^2D}{x-1}+\frac{\left(x-1\right)^2\left(x+1\right)^2x^2F}{x+1}+\frac{\left(x-1\right)^2\left(x+1\right)^2x^2G}{x}$
$x+8=\left(x+1\right)^2x^2A+\left(x-1\right)^2x^2B+\left(x-1\right)^2\left(x+1\right)^2C+\left(x-1\right)\left(x+1\right)^2x^2D+\left(x-1\right)^2\left(x+1\right)x^2F+\left(x-1\right)^2\left(x+1\right)^2xG$
6
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}9=4A&\:\:\:\:\:\:\:(x=1) \\ 7=4B&\:\:\:\:\:\:\:(x=-1) \\ 10=36A+4B+9C+36D+12F+18G&\:\:\:\:\:\:\:(x=2) \\ 6=4A+36B+9C-12D-36F-18G&\:\:\:\:\:\:\:(x=-2) \\ 11=144A+36B+64C+288D+144F+192G&\:\:\:\:\:\:\:(x=3) \\ 5=36A+144B+64C-144D-288F-192G&\:\:\:\:\:\:\:(x=-3)\end{matrix}$
7
Proceed to solve the system of linear equations
$\begin{matrix}4A & + & 0B & + & 0C & + & 0D & + & 0F & + & 0G & =9 \\ 0A & + & 4B & + & 0C & + & 0D & + & 0F & + & 0G & =7 \\ 36A & + & 0B & + & 0C & + & 36D & + & 12F & + & 18G & =10 \\ 0A & + & 36B & + & 9C & - & 12D & - & 36F & - & 18G & =6 \\ 144A & + & 36B & + & 64C & + & 288D & + & 144F & + & 192G & =11 \\ 36A & + & 144B & + & 64C & - & 144D & - & 288F & - & 192G & =5\end{matrix}$
8
Rewrite as a coefficient matrix
$\left(\begin{matrix}4 & 0 & 0 & 0 & 0 & 0 & 9 \\ 0 & 4 & 0 & 0 & 0 & 0 & 7 \\ 36 & 0 & 0 & 36 & 12 & 18 & 10 \\ 0 & 36 & 9 & -12 & -36 & -18 & 6 \\ 144 & 36 & 64 & 288 & 144 & 192 & 11 \\ 36 & 144 & 64 & -144 & -288 & -192 & 5\end{matrix}\right)$
9
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & \frac{9}{4} \\ 0 & 1 & 0 & 0 & 0 & 0 & \frac{7}{4} \\ 0 & 0 & 1 & 0 & 0 & 0 & \frac{32}{37} \\ 0 & 0 & 0 & 1 & 0 & 0 & -2.943243243243243 \\ 0 & 0 & 0 & 0 & 1 & 0 & 2.714414 \\ 0 & 0 & 0 & 0 & 0 & 1 & \frac{49}{370}\end{matrix}\right)$
10
The integral of $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in decomposed fraction equals
$\int\left(\frac{9}{4\left(x-1\right)^2}+\frac{7}{4\left(x+1\right)^2}+\frac{32}{37x^2}+\frac{-2.943243}{x-1}+\frac{2.714414}{x+1}+\frac{\frac{49}{370}}{x}\right)dx$
11
Expand the integral $\int\left(\frac{9}{4\left(x-1\right)^2}+\frac{7}{4\left(x+1\right)^2}+\frac{32}{37x^2}+\frac{-2.943243}{x-1}+\frac{2.714414}{x+1}+\frac{\frac{49}{370}}{x}\right)dx$ into $6$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\frac{9}{4\left(x-1\right)^2}dx+\int\frac{7}{4\left(x+1\right)^2}dx+\int\frac{32}{37x^2}dx+\int\frac{-2.943243}{x-1}dx+\int\frac{2.714414}{x+1}dx+\int\frac{\frac{49}{370}}{x}dx$
12
We can solve the integral $\int\frac{9}{4\left(x-1\right)^2}dx$ 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
$u=x-1$
Intermediate steps
13
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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Intermediate steps
14
Substituting $u$ and $dx$ in the integral and simplify
$\frac{1}{4}\int\frac{9}{u^2}du+\int\frac{7}{4\left(x+1\right)^2}dx+\int\frac{32}{37x^2}dx+\int\frac{-2.943243}{x-1}dx+\int\frac{2.714414}{x+1}dx+\int\frac{\frac{49}{370}}{x}dx$
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Intermediate steps
15
The integral $\frac{1}{4}\int\frac{9}{u^2}du$ results in: $\frac{-9}{4\left(x-1\right)}$
$\frac{-9}{4\left(x-1\right)}$
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Intermediate steps
16
The integral $\int\frac{7}{4\left(x+1\right)^2}dx$ results in: $\frac{-7}{4\left(x+1\right)}$
$\frac{-7}{4\left(x+1\right)}$
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Intermediate steps
17
The integral $\int\frac{32}{37x^2}dx$ results in: $\frac{-32}{37x}$
$\frac{-32}{37x}$
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Intermediate steps
18
The integral $\int\frac{-2.943243}{x-1}dx$ results in: $-2.943243\ln\left(x-1\right)$
$-2.943243\ln\left(x-1\right)$
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Intermediate steps
19
The integral $\int\frac{2.714414}{x+1}dx$ results in: $2.714414\ln\left(x+1\right)$
$2.714414\ln\left(x+1\right)$
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Intermediate steps
20
The integral $\int\frac{\frac{49}{370}}{x}dx$ results in: $\frac{49}{370}\ln\left(x\right)$
$\frac{49}{370}\ln\left(x\right)$
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21
Gather the results of all integrals
$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)$
22
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$
Final Answer
$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$