Final Answer
$\frac{7}{50}\arctan\left(x\right)-\frac{51}{100}\ln\left(x^2+1\right)+\frac{-19}{5\left(x-3\right)}+\frac{51}{50}\ln\left(x-3\right)+C_0$
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Step-by-step Solution
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1
Combining like terms $-x^2$ and $8x^2$
$\int\frac{7x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}dx$
2
Rewrite the fraction $\frac{7x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}$ in $3$ simpler fractions using partial fraction decomposition
$\frac{7x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}$
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 $\left(x^2+1\right)\left(x-3\right)^2$
$7x^2-9x+2=\left(x^2+1\right)\left(x-3\right)^2\left(\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}\right)$
4
Multiplying polynomials
$7x^2-9x+2=\frac{\left(x^2+1\right)\left(x-3\right)^2\left(Ax+B\right)}{x^2+1}+\frac{\left(x^2+1\right)\left(x-3\right)^2C}{\left(x-3\right)^2}+\frac{\left(x^2+1\right)\left(x-3\right)^2D}{x-3}$
$7x^2-9x+2=\left(x-3\right)^2\left(Ax+B\right)+\left(x^2+1\right)C+\left(x^2+1\right)\left(x-3\right)D$
6
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}18=-16A+16B+2C-8D&\:\:\:\:\:\:\:(x=-1) \\ 0=4A+4B+2C-4D&\:\:\:\:\:\:\:(x=1) \\ 38=10C&\:\:\:\:\:\:\:(x=3) \\ 92=-108A+36B+10C-60D&\:\:\:\:\:\:\:(x=-3)\end{matrix}$
7
Proceed to solve the system of linear equations
$\begin{matrix} -16A & + & 16B & + & 2C & - & 8D & =18 \\ 4A & + & 4B & + & 2C & - & 4D & =0 \\ 0A & + & 0B & + & 10C & + & 0D & =38 \\ -108A & + & 36B & + & 10C & - & 60D & =92\end{matrix}$
8
Rewrite as a coefficient matrix
$\left(\begin{matrix}-16 & 16 & 2 & -8 & 18 \\ 4 & 4 & 2 & -4 & 0 \\ 0 & 0 & 10 & 0 & 38 \\ -108 & 36 & 10 & -60 & 92\end{matrix}\right)$
9
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 0 & -\frac{51}{50} \\ 0 & 1 & 0 & 0 & \frac{7}{50} \\ 0 & 0 & 1 & 0 & \frac{19}{5} \\ 0 & 0 & 0 & 1 & \frac{51}{50}\end{matrix}\right)$
10
The integral of $\frac{7x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}$ in decomposed fraction equals
$\int\left(\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}+\frac{19}{5\left(x-3\right)^2}+\frac{51}{50\left(x-3\right)}\right)dx$
11
Expand the integral $\int\left(\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}+\frac{19}{5\left(x-3\right)^2}+\frac{51}{50\left(x-3\right)}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}dx+\int\frac{19}{5\left(x-3\right)^2}dx+\int\frac{51}{50\left(x-3\right)}dx$
Intermediate steps
12
The integral $\int\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}dx$ results in: $-\frac{51}{100}\ln\left(x^2+1\right)+\frac{7}{50}\arctan\left(x\right)$
$-\frac{51}{100}\ln\left(x^2+1\right)+\frac{7}{50}\arctan\left(x\right)$
Explain this step further
Intermediate steps
13
The integral $\int\frac{19}{5\left(x-3\right)^2}dx$ results in: $\frac{-19}{5\left(x-3\right)}$
$\frac{-19}{5\left(x-3\right)}$
Explain this step further
14
Gather the results of all integrals
$\frac{7}{50}\arctan\left(x\right)-\frac{51}{100}\ln\left(x^2+1\right)+\frac{-19}{5\left(x-3\right)}+\int\frac{51}{50\left(x-3\right)}dx$
Intermediate steps
15
The integral $\int\frac{51}{50\left(x-3\right)}dx$ results in: $\frac{51}{50}\ln\left(x-3\right)$
$\frac{51}{50}\ln\left(x-3\right)$
Explain this step further
16
Gather the results of all integrals
$\frac{7}{50}\arctan\left(x\right)-\frac{51}{100}\ln\left(x^2+1\right)+\frac{-19}{5\left(x-3\right)}+\frac{51}{50}\ln\left(x-3\right)$
17
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{7}{50}\arctan\left(x\right)-\frac{51}{100}\ln\left(x^2+1\right)+\frac{-19}{5\left(x-3\right)}+\frac{51}{50}\ln\left(x-3\right)+C_0$
Final Answer$\frac{7}{50}\arctan\left(x\right)-\frac{51}{100}\ln\left(x^2+1\right)+\frac{-19}{5\left(x-3\right)}+\frac{51}{50}\ln\left(x-3\right)+C_0$