Final answer to the problem
$\frac{-3}{28\left(x+1\right)}+\frac{1}{20\left(x-1\right)}+\frac{3}{334}\ln\left(x-6\right)+\frac{5}{98}\ln\left(x+1\right)-\frac{3}{50}\ln\left(x-1\right)+C_0$
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Step-by-step Solution
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Intermediate steps
1
Factor the difference of squares $\left(x^2-1\right)$ as the product of two conjugated binomials
$\int\frac{2x-1}{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)}dx$
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2
Rewrite the fraction $\frac{2x-1}{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)}$ in $5$ simpler fractions using partial fraction decomposition
$\frac{2x-1}{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)}=\frac{A}{\left(x+1\right)^2}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-6}+\frac{D}{x+1}+\frac{F}{x-1}$
3
Find the values for the unknown coefficients: $A, B, C, D, F$. The first step is to multiply both sides of the equation from the previous step by $\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)$
$2x-1=\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)\left(\frac{A}{\left(x+1\right)^2}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-6}+\frac{D}{x+1}+\frac{F}{x-1}\right)$
4
Multiplying polynomials
$2x-1=\frac{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)A}{\left(x+1\right)^2}+\frac{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)B}{\left(x-1\right)^2}+\frac{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)C}{x-6}+\frac{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)D}{x+1}+\frac{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)F}{x-1}$
$2x-1=\left(x-1\right)^2\left(x-6\right)A+\left(x+1\right)^2\left(x-6\right)B+\left(x+1\right)^2\left(x-1\right)^2C+\left(x+1\right)\left(x-1\right)^2\left(x-6\right)D+\left(x+1\right)^2\left(x-1\right)\left(x-6\right)F$
6
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}-3=-28A&\:\:\:\:\:\:\:(x=-1) \\ 1=-20B&\:\:\:\:\:\:\:(x=1) \\ 11=1225C&\:\:\:\:\:\:\:(x=6) \\ -13=-588A-300B+1225C+2940D+2100F&\:\:\:\:\:\:\:(x=-6) \\ 13=36A+64B+2304C+288D+384F&\:\:\:\:\:\:\:(x=7)\end{matrix}$
7
Proceed to solve the system of linear equations
$\begin{matrix} -28A & + & 0B & + & 0C & + & 0D & + & 0F & =-3 \\ 0A & - & 20B & + & 0C & + & 0D & + & 0F & =1 \\ 0A & + & 0B & + & 1225C & + & 0D & + & 0F & =11 \\ -588A & - & 300B & + & 1225C & + & 2940D & + & 2100F & =-13 \\ 36A & + & 64B & + & 2304C & + & 288D & + & 384F & =13\end{matrix}$
8
Rewrite as a coefficient matrix
$\left(\begin{matrix}-28 & 0 & 0 & 0 & 0 & -3 \\ 0 & -20 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1225 & 0 & 0 & 11 \\ -588 & -300 & 1225 & 2940 & 2100 & -13 \\ 36 & 64 & 2304 & 288 & 384 & 13\end{matrix}\right)$
9
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 0 & 0 & \frac{3}{28} \\ 0 & 1 & 0 & 0 & 0 & -\frac{1}{20} \\ 0 & 0 & 1 & 0 & 0 & \frac{3}{334} \\ 0 & 0 & 0 & 1 & 0 & \frac{5}{98} \\ 0 & 0 & 0 & 0 & 1 & -\frac{3}{50}\end{matrix}\right)$
10
The integral of $\frac{2x-1}{\left(x+1\right)^2\left(x-1\right)^2\left(x-6\right)}$ in decomposed fraction equals
$\int\left(\frac{3}{28\left(x+1\right)^2}+\frac{-1}{20\left(x-1\right)^2}+\frac{\frac{3}{334}}{x-6}+\frac{5}{98\left(x+1\right)}+\frac{-3}{50\left(x-1\right)}\right)dx$
11
Expand the integral $\int\left(\frac{3}{28\left(x+1\right)^2}+\frac{-1}{20\left(x-1\right)^2}+\frac{\frac{3}{334}}{x-6}+\frac{5}{98\left(x+1\right)}+\frac{-3}{50\left(x-1\right)}\right)dx$ into $5$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\frac{3}{28\left(x+1\right)^2}dx+\int\frac{-1}{20\left(x-1\right)^2}dx+\int\frac{\frac{3}{334}}{x-6}dx+\int\frac{5}{98\left(x+1\right)}dx+\int\frac{-3}{50\left(x-1\right)}dx$
12
We can solve the integral $\int\frac{3}{28\left(x+1\right)^2}dx$ 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
$\frac{1}{28}\int\frac{3}{u^2}du+\int\frac{-1}{20\left(x-1\right)^2}dx+\int\frac{\frac{3}{334}}{x-6}dx+\int\frac{5}{98\left(x+1\right)}dx+\int\frac{-3}{50\left(x-1\right)}dx$
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Intermediate steps
15
The integral $\frac{1}{28}\int\frac{3}{u^2}du$ results in: $\frac{-3}{28\left(x+1\right)}$
$\frac{-3}{28\left(x+1\right)}$
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Intermediate steps
16
The integral $\int\frac{-1}{20\left(x-1\right)^2}dx$ results in: $\frac{1}{20\left(x-1\right)}$
$\frac{1}{20\left(x-1\right)}$
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Intermediate steps
17
The integral $\int\frac{\frac{3}{334}}{x-6}dx$ results in: $\frac{3}{334}\ln\left(x-6\right)$
$\frac{3}{334}\ln\left(x-6\right)$
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Intermediate steps
18
The integral $\int\frac{5}{98\left(x+1\right)}dx$ results in: $\frac{5}{98}\ln\left(x+1\right)$
$\frac{5}{98}\ln\left(x+1\right)$
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Intermediate steps
19
The integral $\int\frac{-3}{50\left(x-1\right)}dx$ results in: $-\frac{3}{50}\ln\left(x-1\right)$
$-\frac{3}{50}\ln\left(x-1\right)$
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20
Gather the results of all integrals
$\frac{-3}{28\left(x+1\right)}+\frac{1}{20\left(x-1\right)}+\frac{3}{334}\ln\left(x-6\right)+\frac{5}{98}\ln\left(x+1\right)-\frac{3}{50}\ln\left(x-1\right)$
21
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{-3}{28\left(x+1\right)}+\frac{1}{20\left(x-1\right)}+\frac{3}{334}\ln\left(x-6\right)+\frac{5}{98}\ln\left(x+1\right)-\frac{3}{50}\ln\left(x-1\right)+C_0$
Final answer to the problem$\frac{-3}{28\left(x+1\right)}+\frac{1}{20\left(x-1\right)}+\frac{3}{334}\ln\left(x-6\right)+\frac{5}{98}\ln\left(x+1\right)-\frac{3}{50}\ln\left(x-1\right)+C_0$