Question

Evaluate $\int{\dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}}dx}$?

Answer 1

Hint: Assume the given integral as I. Now, multiply the linear terms inside the square root in the denominator and write it as a quadratic equation. Assume this equation as k and differentiate it to find $dk$ in terms of $dx$. Try to write the numerator as the derivative of the quadratic equation and break the integral into two parts and use completing the square method to simplify the quadratic equation. For the first part use the formula $\int{\dfrac{1}{\sqrt{k}}dk}=2\sqrt{k}$ and for the second part use the formula \[\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx}=\ln \left[ x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right]\] to get the answer.

Complete step by step answer:
Here we are asked to find the integral of the function $\dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}}$. Let us assume the given integral as I, so we have,
$\begin{align}
  & \Rightarrow I=\int{\dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}}dx} \\
 & \Rightarrow I=\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}dx} \\
\end{align}$
Assuming ${{x}^{2}}-9x+20=k$ and differentiating both the sides we have,
\[\begin{align}
  & \Rightarrow d\left( {{x}^{2}}-9x+20 \right)=dk \\
 & \Rightarrow \left( 2x-9 \right)dx=dk \\
\end{align}\]
Writing the numerator of the integral of the form of the derivative of the quadratic equation and balancing the terms we get,
$\begin{align}
  & \Rightarrow I=\int{\dfrac{3\left( 2x-9+\dfrac{7}{3}+9 \right)}{\sqrt{{{x}^{2}}-9x+20}}dx} \\
 & \Rightarrow I=\int{\dfrac{3\left( 2x-9 \right)+34}{\sqrt{{{x}^{2}}-9x+20}}dx} \\
\end{align}$
Breaking the integral into two parts we get,
$\begin{align}
  & \Rightarrow I=\int{\dfrac{3\left( 2x-9 \right)}{\sqrt{{{x}^{2}}-9x+20}}dx}+\int{\dfrac{34}{\sqrt{{{x}^{2}}-9x+20}}dx} \\
 & \Rightarrow I=\left[ 3\times \int{\dfrac{\left( 2x-9 \right)}{\sqrt{{{x}^{2}}-9x+20}}dx} \right]+\left[ 34\times \int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}dx} \right] \\
\end{align}$
Now, in the first part we have ${{x}^{2}}-9x+20=k$ and \[\left( 2x-9 \right)dx=dk\]. In the second part using completing the square method given as $a{{x}^{2}}+bx+c=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$ to simplify the quadratic expression we can write ${{x}^{2}}-9x+20={{\left( x-\dfrac{9}{2} \right)}^{2}}-\dfrac{1}{4}$. Therefore the integral becomes,
$\begin{align}
  & \Rightarrow I=\left[ 3\times \int{\dfrac{dk}{\sqrt{k}}} \right]+\left[ 34\times \int{\dfrac{1}{\sqrt{{{\left( x-\dfrac{9}{2} \right)}^{2}}-\dfrac{1}{4}}}dx} \right] \\
 & \Rightarrow I=\left[ 3\times \int{\dfrac{dk}{\sqrt{k}}} \right]+\left[ 34\times \int{\dfrac{1}{\sqrt{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}dx} \right] \\
\end{align}$
Using the basic formulas of integration given as $\int{\dfrac{1}{\sqrt{k}}dk}=2\sqrt{k}$ and \[\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx}=\ln \left[ x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right]\] for the first and second part respectively we get,
\[\Rightarrow I=\left[ 3\times 2\sqrt{k} \right]+\left[ 34\times \ln \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}} \right] \right]\]
Substituting back the value of k and simplifying the expression we get,
\[\therefore I=6\sqrt{{{x}^{2}}-9x+20}+34\ln \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}} \right]+C\]
Here C is the constant of integration as we are evaluating an indefinite integral.

Note: Remember the formulas of integrals of the basic functions which are provided in the list before starting to solve the integral questions. Also remember the differentiation formulas as it will help in the questions which involve substitution methods in the solution of integrals. Here it will be difficult to use the partial fraction method as the denominator involves square root sign.