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Question: Integrate the following question. \(\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\left( {\sqrt...

Integrate the following question.
∫dx(1+x)(x−x2) is equal to\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\left( {\sqrt {x - {x^2}} } \right)}}{\text{ is equal to}}}
A. 1+x(1−x)2+c B. 1+x(1+x)2+c C. 1−x(1−x)2+c D. 2(x−1)1−x+c  {\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 \\\

Explanation

Solution

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

Complete step-by-step solution -
∵x=sin⁡θ\because \sqrt x = \sin \theta
On differentiating we get
⇒12xdx=cos⁡θdθ\Rightarrow \dfrac{1}{{2\sqrt x }}dx = \cos \theta d\theta
⇒dx=2xcos⁡θdθ\Rightarrow dx = 2\sqrt x \cos \theta d\theta (∵x=sin⁡θ)\left( {\because \sqrt x = \sin \theta } \right)
⇒dx=2sin⁡θcos⁡θdθ\Rightarrow dx = 2\sin \theta \cos \theta d\theta
We have to find
∫dx(1+x)(x−x2)\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\left( {\sqrt {x - {x^2}} } \right)}}}
(∵x=sin⁡θ,∴x=sin⁡2θ)\left( {\because \sqrt x = \sin \theta ,\therefore x = {{\sin }^2}\theta } \right)
(x−x2=x(1−x)=sin⁡2θ(1−sin⁡2θ)=sin⁡θ.cos⁡θ)\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 x\sqrt x we get

⇒∫2sin⁡θcos⁡θdθ(1+sin⁡θ)sin⁡θ.cos⁡θ=2∫dθ(1+sin⁡θ)\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 (1−sin⁡θ)\left( {1 - \sin \theta } \right)on numerator and denominator both we get
⇒2∫(1−sin⁡θ)(1−sin⁡2θ)dθ=2∫1−sin⁡θcos⁡2θdθ\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 } } (∵1−sin⁡2θ=cos⁡2θ)\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\\} (∵1cos⁡2θ=sec⁡2θ)\left( {\because \dfrac{1}{{{{\cos }^2}\theta }} = {{\sec }^2}\theta } \right)
⇒2(tan⁡θ−sec⁡θ)+c\Rightarrow 2\left( {\tan \theta - \sec \theta } \right) + c (∵∫sec⁡2θdθ=tan⁡θ)(∵∫sec⁡θ.tan⁡θdθ=tan⁡θ)\left( {\because \int {{{\sec }^2}\theta d\theta = \tan \theta } } \right)\left( {\because \int {\sec \theta .\tan \theta d\theta = \tan \theta } } \right)
=2(x1−x−11−x)+c= 2\left( {\sqrt {\dfrac{x}{{1 - x}}} - \dfrac{1}{{\sqrt {1 - x} }}} \right) + c (∵sin⁡x=x,∴tan⁡x=x1−x,∴sec⁡x=11−x)\left( {\because \sin x = \sqrt x ,\therefore \tan x = \dfrac{{\sqrt x }}{{\sqrt {1 - x} }},\therefore \sec x = \dfrac{1}{{\sqrt {1 - x} }}} \right)
=2(x−1)1−x+c\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.