# Integrate 1/(-6x+5+x^2) from 2 to 4

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

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

## Step by step solution

Problem

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

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

$\begin{matrix}\left(-5\right)\left(-1\right)=5\\ \left(-5\right)+\left(-1\right)=-6\end{matrix}$
2

Thus

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

Using partial fraction decomposition, the fraction $\frac{1}{\left(x-1\right)\left(x-5\right)}$ can be rewritten as

$\frac{1}{\left(x-1\right)\left(x-5\right)}=\frac{A}{x-1}+\frac{B}{x-5}$
4

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

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

Multiplying polynomials

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

Simplifying

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

Expand the polynomial

$1=-5A+Ax-B+Bx$
8

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

$\begin{matrix}1=-2B-6A&\:\:\:\:\:\:\:(x=-1) \\ 1=-4A&\:\:\:\:\:\:\:(x=1)\end{matrix}$
9

Proceed to solve the system of linear equations

$\begin{matrix} -6A & - & 2B & =1 \\ -4A & + & 0B & =1\end{matrix}$
10

Rewrite as a coefficient matrix

$\left(\begin{matrix}-6 & -2 & 1 \\ -4 & 0 & 1\end{matrix}\right)$
11

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

The decomposed integral equivalent is

$\int_{2}^{4}\left(\frac{-\frac{1}{4}}{x-1}+\frac{\frac{1}{4}}{x-5}\right)dx$
13

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

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

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|, where b=-1 and n=-\frac{1}{4} \left[-\frac{1}{4}\ln\left|x-1\right|\right]_{2}^{4}+\int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx 15 Evaluate the definite integral \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx-1\cdot \ln\left|2-1\right|\left(-\frac{1}{4}\right)+\ln\left|4-1\right|\left(-\frac{1}{4}\right) 16 Subtract the values 4 and -1 \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx-1\cdot \ln\left|1\right|\left(-\frac{1}{4}\right)+\ln\left|3\right|\left(-\frac{1}{4}\right) 17 Multiply -\frac{1}{4} times -1 \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx+\ln\left|1\right|\cdot \frac{1}{4}+\ln\left|3\right|\left(-\frac{1}{4}\right) 18 Calculating the absolute value of 3 \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx+\ln\left(1\right)\cdot \frac{1}{4}+\ln\left(3\right)\left(-\frac{1}{4}\right) 19 Calculating the natural logarithm of 3 \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx+0\cdot \frac{1}{4}+\frac{\sqrt[3]{23}}{2}\left(-\frac{1}{4}\right) 20 Any expression multiplied by 0 is equal to 0 \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx+0+\frac{\sqrt[3]{23}}{2}\left(-\frac{1}{4}\right) 21 Multiply -\frac{1}{4} times \frac{\sqrt[3]{23}}{2} \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx+0-\frac{39}{142} 22 x+0=x, where x is any expression \int_{2}^{4}\frac{\frac{1}{4}}{x-5}dx-\frac{39}{142} 23 Apply the formula: \int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=-5$ and $n=\frac{1}{4}$

$\left[\frac{1}{4}\ln\left|x-5\right|\right]_{2}^{4}-\frac{39}{142}$
24

Evaluate the definite integral

$-0.2747-1\cdot \ln\left|2-5\right|\cdot 0.25+\ln\left|4-5\right|\cdot 0.25$
25

Subtract the values $4$ and $-5$

$-0.2747-1\cdot \ln\left|-3\right|\cdot 0.25+\ln\left|-1\right|\cdot 0.25$
26

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

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

Calculating the absolute value of $-1$

$-0.2747+\ln\left(3\right)\left(-0.25\right)+\ln\left(1\right)\cdot 0.25$
28

Calculating the natural logarithm of $1$

$-0.2747+1.0986\left(-0.25\right)+0\cdot 0.25$
29

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

$-0.2747+1.0986\left(-0.25\right)+0$
30

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

$1.0986\left(-0.25\right)-0.2747$
31

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

$-0.2747-0.2747$
32

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

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

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

### Main topic:

Integrals by partial fraction expansion

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