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1

Solved example of base change formula of logarithms

$\int_{2}^{4}\frac{1}{x^2-6x+5}dx$
2

Factor the trinomial $x^2-6x+5$ finding two numbers that multiply to form $5$ and added form $-6$

$\begin{matrix}\left(-1\right)\left(-5\right)=5\\ \left(-1\right)+\left(-5\right)=-6\end{matrix}$
3

Thus

$\int_{2}^{4}\frac{1}{\left(x-1\right)\left(x-5\right)}dx$
4

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

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

Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x-1\right)\left(x-5\right)$

$1=\left(x-1\right)\left(x-5\right)\left(\frac{A}{x-1}+\frac{B}{x-5}\right)$
6

Multiplying polynomials

$1=\frac{A\left(x-1\right)\left(x-5\right)}{x-1}+\frac{B\left(x-1\right)\left(x-5\right)}{x-5}$
7

Simplifying

$1=A\left(x-5\right)+B\left(x-1\right)$
8

Expand the polynomial

$1=Ax-5A+Bx-B$
9

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

$\begin{matrix}1=-6A-2B&\:\:\:\:\:\:\:(x=-1) \\ 1=-4A&\:\:\:\:\:\:\:(x=1)\end{matrix}$
10

Proceed to solve the system of linear equations

$\begin{matrix} -6A & - & 2B & =1 \\ -4A & + & 0B & =1\end{matrix}$
11

Rewrite as a coefficient matrix

$\left(\begin{matrix}-6 & -2 & 1 \\ -4 & 0 & 1\end{matrix}\right)$
12

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{4} \\ 0 & 1 & \frac{1}{4}\end{matrix}\right)$
13

The integral of $\frac{1}{\left(x-1\right)\left(x-5\right)}$ in decomposed fraction equals

$\int_{2}^{4}\left(\frac{-\frac{1}{4}}{x-1}+\frac{\frac{1}{4}}{x-5}\right)dx$
14

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{2}^{4}\frac{-\frac{1}{4}}{x-1}dx+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
15

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-1$ and $n=-\frac{1}{4}$

$\left[-\frac{1}{4}\ln\left|x-1\right|\right]_{2}^{4}+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
16

Evaluate the definite integral

$-\frac{1}{4}\ln\left|4-1\right|-1-\frac{1}{4}\ln\left|2-1\right|+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
17

Subtract the values $2$ and $-1$

$-\frac{1}{4}\ln\left|3\right|-1-\frac{1}{4}\ln\left|1\right|+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
18

Multiply $-1$ times $-\frac{1}{4}$

$-\frac{1}{4}\ln\left|3\right|+\frac{1}{4}\ln\left|1\right|+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
19

Calculating the absolute value of $1$

$-\frac{1}{4}\ln\left(3\right)+\frac{1}{4}\ln\left(1\right)+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
20

Calculating the natural logarithm of $1$

$-\frac{1}{4}\cdot \frac{23}{\sqrt[3]{2}}+\frac{1}{4}\cdot 0+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
21

Any expression multiplied by $0$ is equal to $0$

$-\frac{1}{4}\cdot \frac{23}{\sqrt[3]{2}}+0+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
22

Multiply $-\frac{1}{4}$ times $\frac{23}{\sqrt[3]{2}}$

$-\frac{39}{142}+0+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
23

$x+0=x$, where $x$ is any expression

$-\frac{39}{142}+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx$
24

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-5$ and $n=\frac{1}{4}$

$-\frac{39}{142}+\left[\frac{1}{4}\ln\left|x-5\right|\right]_{2}^{4}$
25

Evaluate the definite integral

$-0.2747+0.25\ln\left|4-5\right|-1\cdot 0.25\ln\left|2-5\right|$
26

Subtract the values $2$ and $-5$

$-0.2747+0.25\ln\left|-1\right|-1\cdot 0.25\ln\left|-3\right|$
27

Multiply $-1$ times $\frac{1}{4}$

$-0.2747+0.25\ln\left|-1\right|-0.25\ln\left|-3\right|$
28

Calculating the absolute value of $-3$

$-0.2747+0.25\ln\left(1\right)-0.25\ln\left(3\right)$
29

Calculating the natural logarithm of $3$

$-0.2747+0.25\cdot 0-0.25\cdot 1.0986$
30

Any expression multiplied by $0$ is equal to $0$

$-0.2747+0-0.25\cdot 1.0986$
31

Subtract the values $0$ and $-\frac{39}{142}$

$-0.2747-0.25\cdot 1.0986$
32

Multiply $-\frac{1}{4}$ times $\frac{23}{\sqrt[3]{2}}$

$-0.2747-0.2747$
33

Subtract the values $-\frac{39}{142}$ and $-\frac{39}{142}$

$-\frac{7}{\sqrt[3]{3}}$

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