Question

Let \[f\left( x \right) = \int {\dfrac{{\sqrt x }}{{{{\left( {1 + x} \right)}^2}}}dx} \], \[x \ge 0\]. Then find the value of \[f\left( 3 \right) - f\left( 1 \right)\].
A. \[ - \dfrac{\pi }{6} + \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}\]
B. \[\dfrac{\pi }{6} + \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{4}\]
C. \[ - \dfrac{\pi }{{12}} + \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}\]
D. \[\dfrac{\pi }{{12}} + \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{4}\]

Answer 1

First we will assume \[x = {\tan ^2}\theta \] and find the derivative of it. Then rewrite the integration as in terms of \[\theta \]. Then simplify the integration. Then we will find the value of \[\theta \] when \[x = 3\] and \[x = 1\] by using the equation \[x = {\tan ^2}\theta \]. After we will apply inverse formula of \[\int\limits_a^b {f\left( x \right)dx} = \phi \left( b \right) - \phi \left( a \right)\] where \[\int {f\left( x \right)dx} = \phi \left( x \right) + C\].

Complete step by step solution
Given integration is
\[f\left( x \right) = \int {\dfrac{{\sqrt x }}{{{{\left( {1 + x} \right)}^2}}}dx} \]
Let \[x = {\tan ^2}\theta \]
Differentiate with respect to \[x\]
\[dx = 2\tan \theta {\sec ^2}\theta d\theta \]
Substitute \[x = {\tan ^2}\theta \] and \[dx = 2\tan \theta {\sec ^2}\theta d\theta \] in the integration \[f\left( x \right) = \int {\dfrac{{\sqrt x }}{{{{\left( {1 + x} \right)}^2}}}dx} \].
\[f\left( \theta \right) = \int {\dfrac{{\sqrt {{{\tan }^2}\theta } }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}2\tan \theta {{\sec }^2}\theta d\theta } \]
Now substitute \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
\[f\left( \theta \right) = \int {\dfrac{{\tan \theta }}{{{{\sec }^4}\theta }}2\tan \theta {{\sec }^2}\theta d\theta } \]
 \[f\left( \theta \right) = 2\int {\dfrac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}d\theta } \]
Now putting \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\sec \theta = \dfrac{1}{{\cos \theta }}\]
\[f\left( \theta \right) = 2\int {\dfrac{{\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{1}{{{{\cos }^2}\theta }}}}d\theta } \]
\[f\left( \theta \right) = 2\int {{{\sin }^2}\theta d\theta } \]
\[f\left( \theta \right) = \int {\left( {1 - \cos 2\theta } \right)d\theta } \]
Now putting \[x = 3\] in the equation \[x = {\tan ^2}\theta \].
\[3 = {\tan ^2}\theta \]
Taking square root both sides
\[\sqrt 3 = \tan \theta \]
\[\theta = \dfrac{\pi }{3}\]
Now putting \[x = 1\] in the equation \[x = {\tan ^2}\theta \].
\[1 = {\tan ^2}\theta \]
Taking square root both sides
\[1 = \tan \theta \]
\[\theta = \dfrac{\pi }{4}\]
Now we will apply the reverse formula \[\int\limits_a^b {f\left( x \right)dx} = \phi \left( b \right) - \phi \left( a \right)\].
\[f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\left( {1 - \cos 2\theta } \right)d\theta } \]
   \[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \left[ {\theta - \dfrac{{\sin 2\theta }}{2}} \right]_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}}\]
\[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \left[ {\dfrac{\pi }{3} - \dfrac{{\sin 2 \cdot \dfrac{\pi }{3}}}{2}} \right] - \left[ {\dfrac{\pi }{4} - \dfrac{{\sin 2 \cdot \dfrac{\pi }{4}}}{2}} \right]\]
\[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \left[ {\dfrac{\pi }{3} - \dfrac{{\sin \dfrac{{2\pi }}{3}}}{2}} \right] - \left[ {\dfrac{\pi }{4} - \dfrac{{\sin \dfrac{\pi }{2}}}{2}} \right]\]
Putting \[\sin \dfrac{{2\pi }}{3} = \dfrac{{\sqrt 3 }}{2}\] and \[\sin \dfrac{\pi }{2} = 1\]
\[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{4} - \dfrac{\pi }{4} + \dfrac{1}{2}\]
\[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \dfrac{{4\pi - 3\pi }}{{12}} - \dfrac{{\sqrt 3 }}{4} + \dfrac{1}{2}\]
\[ \Rightarrow f\left( {\dfrac{\pi }{3}} \right) - f\left( {\dfrac{\pi }{4}} \right) = \dfrac{\pi }{{12}} - \dfrac{{\sqrt 3 }}{4} + \dfrac{1}{2}\]
So the value of \[f\left( 3 \right) - f\left( 1 \right)\] is \[\dfrac{\pi }{{12}} + \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{4}\].
Hence option D is the correct option.

Note: Students often make common mistakes when they start to solve the integration without using a substitute method. They calculate the value \[f\left( 3 \right) - f\left( 1 \right)\] by substituting \[x = 3\] and \[x = 1\] in the integration.