Question

Let \[ f(x)=\int \frac{x\sqrt{x}}{(1+x)^2}\,dx \qquad (x\ge0) \] Then $f(3)-f(1)$ is equal to:

• A. \(\dfrac{-1}{12}+\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}\)
• B. \(\dfrac{1}{12}+\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\)
• C. \(\dfrac{-1}{6}+\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}\)
• D. \(\dfrac{1}{6}+\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\)

Hint:

For integrals involving \(\sqrt{x}\), substitution \(x=t^2\) usually simplifies the expression.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires evaluating the difference between values of an indefinite integral at two points, which is essentially evaluating a definite integral:
\[ f(3) - f(1) = \int_1^3 \frac{\sqrt{x}}{(1+x)^2} dx \]
To solve this, we use substitution. Since we have \(\sqrt{x}\) and \((1+x)\), a trigonometric substitution like \(x = \tan^2 \theta\) is highly effective.
Step 2: Key Formula or Approach:
Let \(x = \tan^2 \theta\).
Then \(dx = 2 \tan \theta \sec^2 \theta d\theta\).
Also, \(\sqrt{x} = \tan \theta\).
And \(1 + x = 1 + \tan^2 \theta = \sec^2 \theta\).
Change of limits:
When \(x = 1\), \(\tan^2 \theta = 1 \implies \theta = \frac{\pi}{4}\).
When \(x = 3\), \(\tan^2 \theta = 3 \implies \theta = \frac{\pi}{3}\).
Step 3: Detailed Explanation:
Substitute everything into the integral:
\[ I = \int_{\pi/4}^{\pi/3} \frac{\tan \theta}{(\sec^2 \theta)^2} \cdot (2 \tan \theta \sec^2 \theta) d\theta \]
\[ I = \int_{\pi/4}^{\pi/3} \frac{2 \tan^2 \theta \sec^2 \theta}{\sec^4 \theta} d\theta \]
\[ I = \int_{\pi/4}^{\pi/3} \frac{2 \tan^2 \theta}{\sec^2 \theta} d\theta \]
Since \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\) and \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\):
\[ I = \int_{\pi/4}^{\pi/3} 2 \sin^2 \theta d\theta \]
Using the trigonometric identity \(2 \sin^2 \theta = 1 - \cos 2\theta\):
\[ I = \int_{\pi/4}^{\pi/3} (1 - \cos 2\theta) d\theta \]
Integrating:
\[ I = \left[ \theta - \frac{\sin 2\theta}{2} \right]_{\pi/4}^{\pi/3} \]
Evaluate at limits:
Upper limit (\(\theta = \pi/3\)):
\[ \frac{\pi}{3} - \frac{\sin(2\pi/3)}{2} = \frac{\pi}{3} - \frac{\sqrt{3}/2}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4} \]
Lower limit (\(\theta = \pi/4\)):
\[ \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} = \frac{\pi}{4} - \frac{1}{2} \]
Final Difference:
\[ I = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right) \]
\[ I = \frac{\pi}{3} - \frac{\pi}{4} + \frac{1}{2} - \frac{\sqrt{3}}{4} \]
Find common denominator for the fractions of \(\pi\):
\[ \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]
So, \(I = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}\).
Step 4: Final Answer:
The result is \(\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}\).