1
Here, we show you a step-by-step solved example of matrices. This solution was automatically generated by our smart calculator:
$\int\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}dx$
Passi intermedi
Rewrite the fraction $\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}$ in $4$ simpler fractions using partial fraction decomposition
$\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}=\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-1}+\frac{D}{x+4}$
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 $\left(x-1\right)^2\left(x+4\right)^2$
$1=\left(x-1\right)^2\left(x+4\right)^2\left(\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-1}+\frac{D}{x+4}\right)$
$1=\frac{\left(x-1\right)^2\left(x+4\right)^2A}{\left(x-1\right)^2}+\frac{\left(x-1\right)^2\left(x+4\right)^2B}{\left(x+4\right)^2}+\frac{\left(x-1\right)^2\left(x+4\right)^2C}{x-1}+\frac{\left(x-1\right)^2\left(x+4\right)^2D}{x+4}$
$1=\left(x+4\right)^2A+\left(x-1\right)^2B+\left(x-1\right)\left(x+4\right)^2C+\left(x-1\right)^2\left(x+4\right)D$
Assigning values to $x$ we obtain the following system of equations
$\begin{matrix}1=25A&\:\:\:\:\:\:\:(x=1) \\ 1=9A+4B-18C+12D&\:\:\:\:\:\:\:(x=-1) \\ 1=25B&\:\:\:\:\:\:\:(x=-4) \\ 1=64A+9B+192C+72D&\:\:\:\:\:\:\:(x=4)\end{matrix}$
Proceed to solve the system of linear equations
$\begin{matrix}25A & + & 0B & + & 0C & + & 0D & =1 \\ 9A & + & 4B & - & 18C & + & 12D & =1 \\ 0A & + & 25B & + & 0C & + & 0D & =1 \\ 64A & + & 9B & + & 192C & + & 72D & =1\end{matrix}$
Rewrite as a coefficient matrix
$\left(\begin{matrix}25 & 0 & 0 & 0 & 1 \\ 9 & 4 & -18 & 12 & 1 \\ 0 & 25 & 0 & 0 & 1 \\ 64 & 9 & 192 & 72 & 1\end{matrix}\right)$
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 0 & \frac{1}{25} \\ 0 & 1 & 0 & 0 & \frac{1}{25} \\ 0 & 0 & 1 & 0 & -\frac{2}{125} \\ 0 & 0 & 0 & 1 & \frac{2}{125}\end{matrix}\right)$
The integral of $\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}$ in decomposed fractions equals
$\frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)}$
2
Rewrite the fraction $\frac{1}{\left(x-1\right)^2\left(x+4\right)^2}$ in $4$ simpler fractions using partial fraction decomposition
$\frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)}$
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3
Expand the integral $\int\left(\frac{1}{25\left(x-1\right)^2}+\frac{1}{25\left(x+4\right)^2}+\frac{-2}{125\left(x-1\right)}+\frac{2}{125\left(x+4\right)}\right)dx$ into $4$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\frac{1}{25\left(x-1\right)^2}dx+\int\frac{1}{25\left(x+4\right)^2}dx+\int\frac{-2}{125\left(x-1\right)}dx+\int\frac{2}{125\left(x+4\right)}dx$
Passi intermedi
Take the constant $\frac{1}{25}$ out of the integral
$\frac{1}{25}\int\frac{1}{\left(x-1\right)^2}dx$
Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=-1$, $c=2$ and $n=1$
$\frac{1}{25}\frac{-1}{\left(2-1\right)\left(x-1\right)^{\left(2-1\right)}}$
$\frac{-1}{25\left(x-1\right)}$
4
The integral $\int\frac{1}{25\left(x-1\right)^2}dx$ results in: $\frac{-1}{25\left(x-1\right)}$
$\frac{-1}{25\left(x-1\right)}$
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Passi intermedi
Take the constant $\frac{1}{25}$ out of the integral
$\frac{1}{25}\int\frac{1}{\left(x+4\right)^2}dx$
Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=4$, $c=2$ and $n=1$
$\frac{1}{25}\frac{-1}{\left(2-1\right)\left(x+4\right)^{\left(2-1\right)}}$
$\frac{-1}{25\left(x+4\right)}$
5
The integral $\int\frac{1}{25\left(x+4\right)^2}dx$ results in: $\frac{-1}{25\left(x+4\right)}$
$\frac{-1}{25\left(x+4\right)}$
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Passi intermedi
Take the constant $\frac{1}{125}$ out of the integral
$\frac{1}{125}\int\frac{-2}{x-1}dx$
Solve the integral $\int\frac{-2}{x-1}dx$ applying change of variable
$-2\left(\frac{1}{125}\right)\ln\left|x-1\right|$
Multiply the fraction and term in $-2\left(\frac{1}{125}\right)\ln\left|x-1\right|$
$-\frac{2}{125}\ln\left|x-1\right|$
6
The integral $\int\frac{-2}{125\left(x-1\right)}dx$ results in: $-\frac{2}{125}\ln\left(x-1\right)$
$-\frac{2}{125}\ln\left(x-1\right)$
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Passi intermedi
Take the constant $\frac{1}{125}$ out of the integral
$\frac{1}{125}\int\frac{2}{x+4}dx$
Solve the integral $\int\frac{2}{x+4}dx$ applying change of variable
$2\left(\frac{1}{125}\right)\ln\left|x+4\right|$
Multiply the fraction and term in $2\left(\frac{1}{125}\right)\ln\left|x+4\right|$
$\frac{2}{125}\ln\left|x+4\right|$
7
The integral $\int\frac{2}{125\left(x+4\right)}dx$ results in: $\frac{2}{125}\ln\left(x+4\right)$
$\frac{2}{125}\ln\left(x+4\right)$
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8
Gather the results of all integrals
$\frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right|$
9
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right|+C_0$
Risposta finale al problema
$\frac{-1}{25\left(x-1\right)}+\frac{-1}{25\left(x+4\right)}-\frac{2}{125}\ln\left|x-1\right|+\frac{2}{125}\ln\left|x+4\right|+C_0$