Chapter 5.7, Problem 33E

To determine

To calculate: The value of given integral by making a substitution to express the integrand as a rational function

Answer to Problem 33E

The value of the given integral is $2+2\ln \frac{5}{3}$

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_9^{16}\frac{\sqrt{x}}{x-4}dx}$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

Calculation:

The given integral is,

  ${\displaystyle {\int }_9^{16}\frac{\sqrt{x}}{x-4}dx}$

Assume, $u=\sqrt{x}$ so

  $\begin{array}{l}{u}^2=x\\ d\left(u^2\right)=dx\left[\text{differentiate both sides}\right]\\ 2udu=dx\end{array}$

Now the given integral becomes,

  $\begin{array}{l}{\displaystyle {\int }_9^{16}\frac{u}{u^2-4}2udu}\\ =2{\displaystyle {\int }_9^{16}\frac{u^2}{u^2-4}du}\end{array}$

The integral can be solved as,

  $\begin{array}{l}{\displaystyle {\int }_9^{16}\frac{\sqrt{x}}{x-4}dx}\\ ={\displaystyle {\int }_9^{16}\frac{u}{u^2-4}2udu}\\ =2{\displaystyle {\int }_9^{16}\frac{u^2-4+4}{u^2-4}}du\\ =2{\displaystyle {\int }_9^{16}\left[\frac{u^2-4}{u^2-4}+\frac{4}{u^2-4}\right]}du\\ =2{\displaystyle {\int }_9^{16}\left[1+\frac{4}{u^2-4}\right]du}\end{array}$

Next the rightmost integral term is written as,

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

Here, $A\text{ and }B$ are the two constants. To find the value of $A\text{ and }B$ multiply both sides by $\left(u-2\right)\left(u+2\right)$

  $\begin{array}{l}4=A\left(u+2\right)+B\left(u-2\right)\\ 4=Au+2A+Bu-2B\\ 4=u\left(A+B\right)+2A-2B\end{array}$

The coefficient of $x$ must be equal and the constant terms are all equal. So

  $A+B=0\text{ and }2A-2B=4$

Solving this system of linear equations of $A\text{ and }B$ the value is found as,

  $\begin{array}{l}A+B=0\text{ and }2A-2B=4\\ A+B=0\Rightarrow B=A\\ -2B=4-2A\Rightarrow -2A=4-2A\Rightarrow A=1,B=-1\end{array}$

So, $A=1\text{ and }B=-1$

Now, the integral is formed the term,

  $2{\displaystyle {\int }_9^{16}\left[1+\frac{1}{u-2}-\frac{1}{u+2}\right]}du$

Simplify the integral and substitute $u$ for $\sqrt{x}$

  $\begin{array}{l}{\displaystyle {\int }_9^{16}\frac{\sqrt{x}}{x-4}dx}\\ =2{\displaystyle {\int }_9^{16}\left[1+\frac{1}{u-2}-\frac{1}{u+2}\right]}du\\ =2{\left[u+\ln \left|u-2\right|-\ln \left|u+2\right|\right]}_9^{16}\\ =2{\left[\sqrt{x}+\ln \left|\sqrt{x}-2\right|-\ln \left|\sqrt{x}+2\right|\right]}_0^{16}\\ =2\left(4+\ln \left|4-2\right|-\ln \left|4+2\right|-3-\ln \left|3-2\right|+\ln \left|3+2\right|\right)\\ =2\left(1+\ln \left(2\right)-\ln \left(6\right)-\ln \left(1\right)+\ln \left(5\right)\right)\\ =2+2\ln \frac{10}{6}\\ =2+2\ln \frac{5}{3}\end{array}$

Therefore, the value of the given integral is $2+2\ln \frac{5}{3}$