1
Solved example of Integration by substitution
$\int\:\frac{-x^2+8x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}dx$
2
Adding $-1x^2$ and $8x^2$
$\int\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}dx$
3
Rewrite the fraction $\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}$ in $3$ simpler fractions using partial fraction decomposition
$\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}=\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}$
4
Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x-3\right)^2\left(x^2+1\right)$
$-9x+2+7x^2=\left(x-3\right)^2\left(x^2+1\right)\left(\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}\right)$
5
Multiplying polynomials
$-9x+2+7x^2=\frac{\left(x-3\right)^2\left(x^2+1\right)\left(Ax+B\right)}{x^2+1}+\frac{C\left(x-3\right)^2\left(x^2+1\right)}{\left(x-3\right)^2}+\frac{D\left(x-3\right)^2\left(x^2+1\right)}{x-3}$
$-9x+2+7x^2=\left(x-3\right)^2\left(Ax+B\right)+C\left(x^2+1\right)+D\left(x^2+1\right)\left(x-3\right)$
$-9x+2+7x^2=\left(x-3\right)^2\left(Ax+B\right)+Cx^2+C+\left(x-3\right)\left(Dx^2+D\right)$
8
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}18=16\left(-A+B\right)-8D+2C&\:\:\:\:\:\:\:(x=-1) \\ 0=4\left(A+B\right)-4D+2C&\:\:\:\:\:\:\:(x=1) \\ 92=36\left(-3A+B\right)+9C+C-60D&\:\:\:\:\:\:\:(x=-3) \\ 38=10C&\:\:\:\:\:\:\:(x=3)\end{matrix}$
9
Proceed to solve the system of linear equations
$\begin{matrix} -1A & + & 1B & + & 2C & - & 8D & =18 \\ 1A & + & 1B & + & 2C & - & 4D & =0 \\ -3A & + & 1B & + & 10C & - & 60D & =92 \\ 0A & + & 0B & + & 10C & + & 0D & =38\end{matrix}$
10
Rewrite as a coefficient matrix
$\left(\begin{matrix}-1 & 1 & 2 & -8 & 18 \\ 1 & 1 & 2 & -4 & 0 \\ -3 & 1 & 10 & -60 & 92 \\ 0 & 0 & 10 & 0 & 38\end{matrix}\right)$
11
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 0 & -7.9333 \\ 0 & 1 & 0 & 0 & -\frac{9}{5} \\ 0 & 0 & 1 & 0 & \frac{19}{5} \\ 0 & 0 & 0 & 1 & -\frac{8}{15}\end{matrix}\right)$
12
The decomposed integral equivalent is
$\int\left(\frac{-7.9333x-\frac{9}{5}}{x^2+1}+\frac{\frac{19}{5}}{\left(x-3\right)^2}+\frac{-\frac{8}{15}}{x-3}\right)dx$
13
The integral of a sum of two or more functions is equal to the sum of their integrals
$\int\frac{-7.9333x-\frac{9}{5}}{x^2+1}dx+\int\frac{\frac{19}{5}}{\left(x-3\right)^2}dx+\int\frac{-\frac{8}{15}}{x-3}dx$
14
Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-3$ and $n=-\frac{8}{15}$
$\int\frac{-7.9333x-\frac{9}{5}}{x^2+1}dx+\int\frac{\frac{19}{5}}{\left(x-3\right)^2}dx-\frac{8}{15}\ln\left|x-3\right|$
15
Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(x+a\right)^{\left(c-1\right)}\left(c-1\right)}$, where $a=-3$, $c=2$ and $n=\frac{19}{5}$
$\int\frac{-7.9333x-\frac{9}{5}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
16
Split the fraction $\frac{-7.9333x+-1.8}{x^2+1}$ in two terms with same denominator
$\int\left(\frac{-7.9333x}{x^2+1}+\frac{-\frac{9}{5}}{x^2+1}\right)dx+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
17
The integral of a sum of two or more functions is equal to the sum of their integrals
$\int\frac{-7.9333x}{x^2+1}dx+\int\frac{-\frac{9}{5}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
18
Taking the constant out of the integral
$-7.9333\int\frac{x}{x^2+1}dx+\int\frac{-\frac{9}{5}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
19
Take the constant out of the integral
$-7.9333\int\frac{x}{x^2+1}dx-\frac{9}{5}\int\frac{1}{1+x^2}dx+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
20
Solve the integral applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$
$-7.9333\int\frac{x}{x^2+1}dx-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
21
Solve the integral $\int\frac{x}{x^2+1}dx$ applying u-substitution. Let $u$ and $du$ be
$\begin{matrix}u=x^2+1 \\ du=2xdx\end{matrix}$
22
Isolate $dx$ in the previous equation
$\frac{du}{2x}=dx$
23
Substituting $u$ and $dx$ in the integral and simplify
$-7.9333\int\frac{1}{2u}du-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
24
Take the constant out of the integral
$-3.9667\int\frac{1}{u}du-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
25
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
$-3.9667\ln\left|u\right|-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
26
Substitute $u$ back for it's value, $x^2+1$
$-3.9667\ln\left|x^2+1\right|-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|$
27
As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration
$-3.9667\ln\left|x^2+1\right|-\frac{9}{5}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}-\frac{8}{15}\ln\left|x-3\right|+C_0$