Question

Integrate the following function

\[\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}\]

Answer 1

Hint: In this question, first of all, convert the denominator that is (x – a) (x – b) in the form of \[{{x}^{2}}-{{k}^{2}}\] by using the formulas \[{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\] and \[{{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\]. Then use \[\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{k}^{2}}}}}=\ln \left| x+\sqrt{{{x}^{2}}-{{c}^{2}}} \right|+k\] and simplify it to get the answer.

Complete step-by-step answer:
Here, we have to integrate the function \[\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}\].
Let us consider the integral given in the question.

\[I=\int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}\]

We know that \[\left( x-a \right)\left( x-b \right)={{x}^{2}}-\left( a+b \right)x+ab\]. So, we get,

\[I=\int{\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}}dx\]

Now, let us assume \[{{x}^{2}}-\left( a+b \right)x+ab=D\left( x \right)\]. So, we get,

\[I=\int{\dfrac{1}{\sqrt{D\left( x \right)}}dx....\left( i \right)}\]

Now, let us consider D(x).

\[D\left( x \right)={{x}^{2}}-\left( a+b \right)x+ab\]

Now, we will add and subtract \[{{\left( \dfrac{a+b}{2} \right)}^{2}}\] in RHS of the above equation, we get,

\[D\left( x \right)={{x}^{2}}-\left( a+b \right)x+ab+{{\left( \dfrac{a+b}{2} \right)}^{2}}-{{\left(

\dfrac{a+b}{2} \right)}^{2}}\]

By rearranging the terms of the above equation, we get,

\[D\left( x \right)={{x}^{2}}-2\left( \dfrac{a+b}{2} \right)x+{{\left( \dfrac{a+b}{2}

 \right)}^{2}}+ab-{{\left( \dfrac{a+b}{2} \right)}^{2}}\]
We know that \[{{x}^{2}}+{{y}^{2}}+2xy={{\left( x+y \right)}^{2}}\] and \[{{x}^{2}}+{{y}^{2}}-2xy={{\left( x-y \right)}^{2}}\]. By using these formulas in the above equation, we get,

\[D\left( x \right)={{\left[ x-\left( \dfrac{a+b}{2} \right) \right]}^{2}}+ab-\left[

\dfrac{{{a}^{2}}+{{b}^{2}}+2ab}{4} \right]\]

By simplifying the above equation, we get,

\[D\left( x \right)={{\left[ x-\left( \dfrac{a+b}{2} \right) \right]}^{2}}-\left[

\dfrac{{{a}^{2}}+{{b}^{2}}+2ab-4ab}{4} \right]\]

\[D\left( x \right)={{\left[ x-\left( \dfrac{a+b}{2} \right) \right]}^{2}}-\left(

\dfrac{{{a}^{2}}+{{b}^{2}}-2ab}{4} \right)\]

\[\Rightarrow D\left( x \right)={{\left[ x-\left( \dfrac{a+b}{2} \right) \right]}^{2}}-{{\left(

\dfrac{a-b}{2} \right)}^{2}}\]

By substituting the values of D(x) in equation (i), we get,

\[I=\int{\dfrac{1}{{{\left[ x-\dfrac{\left( a+b \right)}{2} \right]}^{2}}-{{\left( \dfrac{a-b}{2}

\right)}^{2}}}dx}\]

Now, let us take \[\left[ x-\dfrac{\left( a+b \right)}{2} \right]=t\]

So, by differentiating both the sides, we get, dx = dt. So, we get,

\[I=\int{\dfrac{1}{{{t}^{2}}-{{\left( \dfrac{a-b}{2} \right)}^{2}}}}dt\]

We know that, \[\int{\dfrac{1}{{{x}^{2}}-{{c}^{2}}}dx}=\ln \left| x+\sqrt{{{x}^{2}}+{{c}^{2}}}

\right|+k\]

where k and c are constants. So by using this in the above equation, we get,

\[I=\ln \left| t+\sqrt{\left[ {{t}^{2}}-{{\left( \dfrac{a-b}{2} \right)}^{2}} \right]} \right|+k\]

By substituting the value of t in the above equation (i), we get,

\[I=\ln \left| \left( x-\left( \dfrac{a+b}{2} \right) \right)+\sqrt{\left[ x-{{\left( \dfrac{a+b}{2}

\right)}^{2}}-{{\left( \dfrac{a-b}{2} \right)}^{2}} \right]} \right|+k\]

Let us take \[E={{\left( x-\left( \dfrac{a+b}{2} \right) \right)}^{2}}-{{\left( \dfrac{a-b}{2}

 \right)}^{2}}\]

So, we get, \[I=\ln \left| \left( x-\left( \dfrac{a+b}{2} \right) \right)+\sqrt{E} \right|+k....\left( ii

\right)\]

We know that \[{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\] and \[{{\left( x-y

\right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\]

So, by simplifying the expression for E, we get,

\[E={{x}^{2}}+{{\left( \dfrac{a+b}{2} \right)}^{2}}-2.x.\left( \dfrac{a+b}{2} \right)-{{\left(

\dfrac{a-b}{2} \right)}^{2}}\]

\[E={{x}^{2}}-x\left( a+b \right)+{{\left( \dfrac{a+b}{2} \right)}^{2}}-{{\left( \dfrac{a-b}{2}

\right)}^{2}}\]

\[E={{x}^{2}}-x\left( a+b \right)+\left( \dfrac{{{a}^{2}}+{{b}^{2}}+2ab}{4} \right)-\left(

\dfrac{{{a}^{2}}+{{b}^{2}}-2ab}{4} \right)\]

By canceling out the like terms, we get,

\[\Rightarrow E={{x}^{2}}-x\left( a+b \right)+\dfrac{4ab}{4}\]

\[E={{x}^{2}}-x\left( a+b \right)+ab\]

\[\Rightarrow E=\left( x-a \right)\left( x-b \right)\]

By substituting the value of E in equation (ii), we get,

\[I=\ln \left| \left( x-\dfrac{\left( a+b \right)}{2} \right)+\sqrt{\left( x-a \right)\left( x-b \right)}

\right|+k\]

So, we get,

\[\int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\ln \left| \left( x-\left( \dfrac{a+b}{2}

\right) \right)+\sqrt{\left( x-a \right)\left( x-b \right)} \right|+k}\]


Note: In this question, students must take care of the sign of terms while taking common terms and rearranging the terms. As the question involves a lot of algebra, students are advised to check each equation twice to avoid any mistakes because that can lead to the whole solution being incorrect. Also, in these types of questions, students should always try to convert the term into perfect squares by adding and subtracting a constant term.