Integrate 1/((x-1)(x+2)) from 2 to 5

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

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Answer

$\frac{62}{225}$

Step by step solution

Problem

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

Using partial fraction decomposition, the fraction $\frac{1}{\left(2+x\right)\left(x-1\right)}$ can be rewritten as

$\frac{1}{\left(2+x\right)\left(x-1\right)}=\frac{A}{2+x}+\frac{B}{x-1}$
2

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(2+x\right)\left(x-1\right)$

$1=\left(\frac{A}{2+x}+\frac{B}{x-1}\right)\left(2+x\right)\left(x-1\right)$
3

Multiplying polynomials

$1=\frac{A\left(2+x\right)\left(x-1\right)}{2+x}+\frac{B\left(2+x\right)\left(x-1\right)}{x-1}$
4

Simplifying

$1=A\left(x-1\right)+B\left(2+x\right)$
5

Expand the polynomial

$1=-A+Ax+2B+Bx$
6

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

$\begin{matrix}1=B-2A&\:\:\:\:\:\:\:(x=-1) \\ 1=-3A&\:\:\:\:\:\:\:(x=-2)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix} -2A & + & 1B & =1 \\ -3A & + & 0B & =1\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}-2 & 1 & 1 \\ -3 & 0 & 1\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{3} \\ 0 & 1 & \frac{1}{3}\end{matrix}\right)$
10

The decomposed integral equivalent is

$\int_{2}^{5}\left(\frac{-\frac{1}{3}}{2+x}+\frac{\frac{1}{3}}{x-1}\right)dx$
11

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

$\int_{2}^{5}\frac{-\frac{1}{3}}{2+x}dx+\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx$
12

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=2$ and $n=-\frac{1}{3}$

$\left[-\frac{1}{3}\ln\left|2+x\right|\right]_{2}^{5}+\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx$
13

Evaluate the definite integral

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx-1\cdot \ln\left|2+2\right|\left(-\frac{1}{3}\right)+\ln\left|2+5\right|\left(-\frac{1}{3}\right)$
14

Add the values $5$ and $2$

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx-1\cdot \ln\left|4\right|\left(-\frac{1}{3}\right)+\ln\left|7\right|\left(-\frac{1}{3}\right)$
15

Multiply $-\frac{1}{3}$ times $-1$

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx+\ln\left|4\right|\cdot \frac{1}{3}+\ln\left|7\right|\left(-\frac{1}{3}\right)$
16

Calculating the absolute value of $7$

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx+\ln\left(4\right)\cdot \frac{1}{3}+\ln\left(7\right)\left(-\frac{1}{3}\right)$
17

Calculating the natural logarithm of $7$

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx+\frac{\sqrt{17}}{3}\cdot \frac{1}{3}+\sqrt[3]{14}\left(-\frac{1}{3}\right)$
18

Multiply $-\frac{1}{3}$ times $\sqrt[3]{14}$

$\int_{2}^{5}\frac{\frac{1}{3}}{x-1}dx+\frac{\sqrt{3}}{4}-\frac{\sqrt[4]{8}}{2}$
19

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=-1$ and $n=\frac{1}{3}$

$\left[\frac{1}{3}\ln\left|x-1\right|\right]_{2}^{5}+\frac{\sqrt{3}}{4}-\frac{\sqrt[4]{8}}{2}$
20

Evaluate the definite integral

$0.4621-0.6486-1\cdot \ln\left|2-1\right|\cdot 0.3333+\ln\left|5-1\right|\cdot 0.3333$
21

Subtract the values $5$ and $-1$

$-1\cdot \ln\left|1\right|\cdot 0.3333+\ln\left|4\right|\cdot 0.3333-0.1865$
22

Multiply $\frac{1}{3}$ times $-1$

$\ln\left|1\right|\left(-0.3333\right)+\ln\left|4\right|\cdot 0.3333-0.1865$
23

Calculating the absolute value of $4$

$\ln\left(1\right)\left(-0.3333\right)+\ln\left(4\right)\cdot 0.3333-0.1865$
24

Calculating the natural logarithm of $4$

$0\left(-0.3333\right)+1.3863\cdot 0.3333-0.1865$
25

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

$0+1.3863\cdot 0.3333-0.1865$
26

Subtract the values $0$ and $-\frac{61}{327}$

$1.3863\cdot 0.3333-0.1865$
27

Multiply $\frac{1}{3}$ times $\frac{\sqrt{17}}{3}$

$0.4621-0.1865$
28

Subtract the values $\frac{\sqrt{3}}{4}$ and $-\frac{61}{327}$

$\frac{62}{225}$

Answer

$\frac{62}{225}$