Question

${\int }_5^{10}\frac{1}{\left[\left(x-1\right)\left(x-2\right)\right]}dx=$

• A. $\log \left(\frac{27}{32}\right)$
• B. $\log \left(\frac{32}{27}\right)$
• C. $\log \left(\frac{8}{9}\right)$
• D. $\log \left(\frac{3}{4}\right)$

Answer 1

The correct option is B

$\log \left(\frac{32}{27}\right)$

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_5^{10}\frac{1}{\left[\left(x-1\right)\left(x-2\right)\right]}dx\dots \dots \left(i\right)$

Consider $\frac{1}{\left[\left(x-1\right)\left(x-2\right)\right]}=\frac{A}{x-1}+\frac{B}{x-2}$ $\left[\mathrm{integration}\mathrm{by}\mathrm{parts}\right]$

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

$\Rightarrow$ $1=Ax-2A+Bx-B$

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

$\Rightarrow$ $A+B=0\dots \dots \left(ii\right)$

$\Rightarrow$ $-2A-B=1\dots \dots \left(ii\right)$

on solving the equation $\left(ii\right)$ and $\left(iii\right)$ we get

$A=-1$

$B=1$

Therefore $\left(i\right)$ becomes

$\Rightarrow$ $I={\int }_5^{10}\left[\frac{-1}{\left(x-1\right)}+\frac{1}{x-2}\right]dx$

$={\left[-\log \left(x-1\right)+\log \left(x-2\right)\right]}_5^{10}$

$={\left[\log \left(\frac{x-2}{x-1}\right)\right]}_5^{10}$

$=\left[\log \left(\frac{10-2}{10-1}\right)-\log \left(\frac{5-2}{5-1}\right)\right]$

$=\left[\log \left(\frac{8}{9}\right)-\log \left(\frac{3}{4}\right)\right]$

$=\log \left(\frac{32}{27}\right)$

Thus, option B is the correct answer.