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1

Here, we show you a step-by-step solved example of matrices. This solution was automatically generated by our smart calculator:

$\int \frac { 32 x - 20 } { ( x - 1 ) ( 5 x - 3 ) } d x$

Rewrite the fraction $\frac{32x-20}{\left(x-1\right)\left(5x-3\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{32x-20}{\left(x-1\right)\left(5x-3\right)}=\frac{A}{x-1}+\frac{B}{5x-3}$

Find the values for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step by $\left(x-1\right)\left(5x-3\right)$

$32x-20=\left(x-1\right)\left(5x-3\right)\left(\frac{A}{x-1}+\frac{B}{5x-3}\right)$

Multiplying polynomials

$32x-20=\frac{\left(x-1\right)\left(5x-3\right)A}{x-1}+\frac{\left(x-1\right)\left(5x-3\right)B}{5x-3}$

Simplifying

$32x-20=\left(5x-3\right)A+\left(x-1\right)B$

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}12=2A&\:\:\:\:\:\:\:(x=1) \\ -52=-8A-2B&\:\:\:\:\:\:\:(x=-1)\end{matrix}$

Proceed to solve the system of linear equations

$\begin{matrix}2A & + & 0B & =12 \\ -8A & - & 2B & =-52\end{matrix}$

Rewrite as a coefficient matrix

$\left(\begin{matrix}2 & 0 & 12 \\ -8 & -2 & -52\end{matrix}\right)$

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 6 \\ 0 & 1 & 2\end{matrix}\right)$

The integral of $\frac{32x-20}{\left(x-1\right)\left(5x-3\right)}$ in decomposed fractions equals

$\frac{6}{x-1}+\frac{2}{5x-3}$
2

Rewrite the fraction $\frac{32x-20}{\left(x-1\right)\left(5x-3\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{6}{x-1}+\frac{2}{5x-3}$
3

Expand the integral $\int\left(\frac{6}{x-1}+\frac{2}{5x-3}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{6}{x-1}dx+\int\frac{2}{5x-3}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=6$

$6\ln\left|x-1\right|$
4

The integral $\int\frac{6}{x-1}dx$ results in: $6\ln\left(x-1\right)$

$6\ln\left(x-1\right)$

Apply the formula: $\int\frac{n}{a+b}dx$$=n\int\frac{1}{a+b}dx$, where $a=-3$, $b=5x$ and $n=2$

$2\int\frac{1}{-3+5x}dx$

Apply the formula: $\int\frac{n}{ax+b}dx$$=\frac{n}{a}\ln\left(ax+b\right)+C$, where $a=5$, $b=-3$ and $n=1$

$2\cdot \left(\frac{1}{5}\right)\ln\left|5x-3\right|$

Apply the formula: $\frac{a}{b}c$$=\frac{ca}{b}$, where $a=1$, $b=5$, $c=2$, $a/b=\frac{1}{5}$ and $ca/b=2\cdot \left(\frac{1}{5}\right)\ln\left(5x-3\right)$

$\frac{2}{5}\ln\left|5x-3\right|$
5

The integral $\int\frac{2}{5x-3}dx$ results in: $\frac{2}{5}\ln\left(5x-3\right)$

$\frac{2}{5}\ln\left(5x-3\right)$
6

Gather the results of all integrals

$6\ln\left|x-1\right|+\frac{2}{5}\ln\left|5x-3\right|$
7

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$6\ln\left|x-1\right|+\frac{2}{5}\ln\left|5x-3\right|+C_0$

Final answer to the problem

$6\ln\left|x-1\right|+\frac{2}{5}\ln\left|5x-3\right|+C_0$

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