Chapter G, Problem 35E

To determine

To calculate:

The integral value of the following functions by using substitution method:

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}$

Answer to Problem 35E

The integral value of ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}$ is $2+\ln \frac{25}{9}$ .

Explanation of Solution

Given information:

The integral function:

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}$

Calculation:

The integral function is given that:

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}$

Substitute the value:

  $\begin{array}{l}u=\sqrt{x}\\ du=\frac{1}{2\sqrt{x}}dx\\ 2udu=dx\end{array}$

Limits are changed:

If $x=9$ then, $u=3$

If $x=16$ then, $u=4$

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}={\displaystyle \underset{3}{\overset{4}{\int }}\frac{2u^2}{u^2-4}du}$.........(1)

Firstly take the integrand and then decompose using the partial fractions:

  $\frac{u^2}{u^2-4}=1+\frac{4}{u^2-4}$

  $\frac{u^2}{u^2-4}=1+\frac{4}{\left(u-2\right)\left(u+2\right)}$

Use the partial fraction:

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

Equate numerators:

  $4=A\left(u+2\right)+B(u-2)$

Put $u=2$ , then $4=4A+0,$ that implies $A=1$ .

Put $u=-2$ , then $4=0-4B,$ that implies $B=-1$ .

  $\frac{4}{\left(u-2\right)\left(u+2\right)}=\frac{1}{u-2}+\frac{1}{u+2}$

  $\frac{u^2}{u^2-4}=1+\frac{1}{\left(u-2\right)}+\frac{1}{\left(u+2\right)}$

Integral equation (1) can be written as,

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}={\displaystyle \underset{3}{\overset{4}{\int }}\frac{2u^2}{u^2-4}du}$

  ${\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2{\displaystyle \underset{3}{\overset{4}{\int }}\left(1+\frac{1}{u-2}+\frac{1}{u+2}\right)du}$

  $\begin{array}{l}{\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2{\left[u+\ln \left(u-2\right)-\ln \left(u+2\right)\right]}_3^4\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2{\left[u+\ln \frac{u-2}{u+2}\right]}_3^4\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2{\left[4+\ln \frac{2}{6}-3-\ln \frac{1}{5}\right]}_3^4\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=8+2\ln \frac{1}{3}-6-2\ln \frac{1}{5}\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2+2\ln \frac{5}{3}\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2+\ln {\left(\frac{5}{3}\right)}^2\\ {\displaystyle \underset{9}{\overset{16}{\int }}\frac{\sqrt{x}}{x-4}dx}=2+\ln \frac{25}{9}\end{array}$