Question

Make a substitution to express the integrated as a rational fraction and then evaluate the integral.

\(\int\limits_9^{16} {\frac{{\sqrt x }}{{x - 4}}dx} \) ( let \(u = \sqrt x \)\(\))

Short answer

The solution of the given integral is given as \(\)\(2 + \ln \frac{{25}}{9}\)

Step-by-step solution

Step-1: Rewrite the integral and use partial fraction

Let\(u = \sqrt x \) \(du = \frac{1}{{2\sqrt x }}dx\) or \(2udu = dx\)

Limits are changed as follows: - if x=9, then u=3

If x=16 ,then u=4

Rewrite the integral as \(\int\limits_9^{16} {\frac{{\sqrt x }}{{x - 4}}dx} = \int\limits_3^4 {\frac{{2{u^2}}}{{{u^2} + 4}}} du\)

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

= \(\)\(1 + \frac{4}{{\left( {u - 2} \right)\left( {u - 2} \right)}}\) \(\left( {because,{a^2} - {b^2} = (a + b)\left( {a - b} \right)} \right)\)

Use partial function s to write,

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

Step-2:- Find the value of A and B

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

Therefore,\(4 = \left( {u + 2} \right)A + B\left( {u - 2} \right) = > 4 = Au + 2A + Bu - 2B\)

\(2A - 2B = 4\)----(1) and\(A + B = 0 - - - \left( 2 \right)\) (compare the like terms)

Add both (1) and (2),

\(\begin{array}{l}2A - 2B = 4 = > A - B = 2\\A - B + A + B = 2 + 0 = > 2A = 2 = > A = 1\end{array}\)

Substitute A=1 in equation (2), we get 1+B=> B=-1

= \(\frac{4}{{\left( {u - 2} \right)\left( {u + 2} \right)}} = \frac{1}{{u - 2}} + \frac{1}{{u + 2}} - - - \left( 3 \right)\)

Step-(3):- Find the integral

= \(\int\limits_9^{16} {\frac{{\sqrt x }}{{x - 4}}dx} = \int\limits_3^4 {\frac{{2{u^2}}}{{{u^2} + 4}}} du\)=\(2\int\limits_3^4 {1 + \frac{4}{{\left( {u + 2} \right)\left( {u - 2} \right)}}du} \)

= \(2\int\limits_3^4 {1 + \frac{4}{{\left( {u + 2} \right)\left( {u - 2} \right)}}du} \) ( from equation (3))

=\(2(u + \ln \left( {u - 2} \right) - \ln \left( {u + 2} \right))_3^4\) \(\left( {\int {\frac{{{f^'}\left( x \right)}}{{f\left( x \right)}}} } \right)dx = \ln \left( {f\left( x \right)} \right) + c\))

= \(2(u + \ln \frac{{\left( {u - 2} \right)}}{{\left( {u + 2} \right)}})_3^4\) \(\left( {\ln \frac{a}{b} = \ln a - \ln b} \right)\)

= \(2(4 + \ln \frac{2}{6} - 3 - \ln \frac{1}{5})\)

=\(8 + 2\ln \frac{1}{3} - 6 - 2\ln \frac{1}{5} = 2 + 2\ln \frac{5}{3}\)

=\(2 + 2\ln {(\frac{5}{3})^2} = 2 + \ln \left( {\frac{{25}}{9}} \right)\)

Hence, the value of the given integral is given below:

\(\int\limits_9^{16} {\frac{{\sqrt x }}{{x - 4}}dx} = 2 + \ln \left( {\frac{{25}}{9}} \right)\)