Question

Integrate the following question.
$\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\left( {\sqrt {x - {x^2}} } \right)}}{\text{ is equal to}}} $
$
  {\text{A}}{\text{. }}\dfrac{{1 + \sqrt x }}{{{{\left( {1 - x} \right)}^2}}} + c \\
  {\text{B}}{\text{. }}\dfrac{{1 + \sqrt x }}{{{{\left( {1 + x} \right)}^2}}} + c \\
  {\text{C}}{\text{. }}\dfrac{{1 - \sqrt x }}{{{{\left( {1 - x} \right)}^2}}} + c \\
  {\text{D}}{\text{. }}\dfrac{{2\left( {\sqrt x - 1} \right)}}{{\sqrt {1 - x} }} + c \\
 $

Answer 1

Hint: When you see this type of question it seems very complex so to make it easy use the substitution method put \[\sqrt x = \sin \theta \] and solve further as a simple integration question.

Complete step-by-step solution -
$\because \sqrt x = \sin \theta $
On differentiating we get
$ \Rightarrow \dfrac{1}{{2\sqrt x }}dx = \cos \theta d\theta $
$ \Rightarrow dx = 2\sqrt x \cos \theta d\theta $$\left( {\because \sqrt x = \sin \theta } \right)$
$ \Rightarrow dx = 2\sin \theta \cos \theta d\theta $
We have to find
$\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\left( {\sqrt {x - {x^2}} } \right)}}} $
$\left( {\because \sqrt x = \sin \theta ,\therefore x = {{\sin }^2}\theta } \right)$
$\left( {\sqrt {x - {x^2}} = \sqrt {x\left( {1 - x} \right)} = \sqrt {{{\sin }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)} = \sin \theta .\cos \theta } \right)$
On putting the values of dx and $\sqrt x $ we get

$ \Rightarrow \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sin \theta .\cos \theta }}} = 2\int {\dfrac{{d\theta }}{{\left( {1 + \sin \theta } \right)}}} $
On multiplying by $\left( {1 - \sin \theta } \right)$on numerator and denominator both we get
$ \Rightarrow 2\int {\dfrac{{\left( {1 - \sin \theta } \right)}}{{\left( {1 - {{\sin }^2}\theta } \right)}}d\theta = 2\int {\dfrac{{1 - \sin \theta }}{{{{\cos }^2}\theta }}d\theta } } $$\left( {\because 1 - {{\sin }^2}\theta = {{\cos }^2}\theta } \right)$
$ \Rightarrow \left\{ {\int {{{\sec }^2}\theta d\theta - \int {\left( {\tan \theta .\sec \theta } \right)d\theta } } } \right\}$$\left( {\because \dfrac{1}{{{{\cos }^2}\theta }} = {{\sec }^2}\theta } \right)$
$ \Rightarrow 2\left( {\tan \theta - \sec \theta } \right) + c$ $\left( {\because \int {{{\sec }^2}\theta d\theta = \tan \theta } } \right)\left( {\because \int {\sec \theta .\tan \theta d\theta = \tan \theta } } \right)$
$ = 2\left( {\sqrt {\dfrac{x}{{1 - x}}} - \dfrac{1}{{\sqrt {1 - x} }}} \right) + c$$\left( {\because \sin x = \sqrt x ,\therefore \tan x = \dfrac{{\sqrt x }}{{\sqrt {1 - x} }},\therefore \sec x = \dfrac{1}{{\sqrt {1 - x} }}} \right)$
=$\dfrac{{2\left( {\sqrt x - 1} \right)}}{{\sqrt {1 - x} }} + c$
Hence option D is the correct option.

Note: These are some special types of questions which are easily solved by substitution method. If you solve it by usual method it will go very long or it will not be solved. So proceed from the substitution method and use simple integration and simple trigonometric results to get an answer.