Question

Let \[ f(x)=\int \frac{\sqrt{x}}{(1+x)^2}\,dx \quad (x\geq 0) \] Then \[ f(3)-f(1) \] is equal to:

• A. $ -\frac{\pi}{12} + \frac{1}{2} + \frac{\sqrt{3}}{4} $
• B. $ \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} $
• C. $ -\frac{\pi}{6} + \frac{1}{2} + \frac{\sqrt{3}}{4} $
• D. $ \frac{\pi}{6} + \frac{1}{2} - \frac{\sqrt{3}}{4} $

Hint:

When you see $ \sqrt{x} $ and $ (1+x) $ in the same integral, the substitution $ x = \tan^2 \theta $ is usually the fastest way to collapse the denominator using Pythagorean identities.

Answer 1

The Correct Option is B

Concept: The expression $ f(3) - f(1) $ represents the definite integral of the function from 1 to 3. Specifically, by the Fundamental Theorem of Calculus: $$ f(3) - f(1) = \int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} dx $$ To solve this, we use a trigonometric substitution to simplify the radical and the denominator. Since we have a term like $ (1+x) $, substituting $ x = \tan^2 \theta $ is effective because $ 1 + \tan^2 \theta = \sec^2 \theta $.

Step 1: Substitution and changing limits. Let $ x = \tan^2 \theta $. Then: $$ dx = 2 \tan \theta \sec^2 \theta \, d\theta $$ Now, let's find the new limits for $ \theta $:
• When $ x = 1 $, $ \tan^2 \theta = 1 \Rightarrow \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4} $.
• When $ x = 3 $, $ \tan^2 \theta = 3 \Rightarrow \tan \theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3} $.

Step 2: Transforming the integral. Substitute the values into the integral: $$ I = \int_{\pi/4}^{\pi/3} \frac{\sqrt{\tan^2 \theta}}{(1 + \tan^2 \theta)^2} \cdot (2 \tan \theta \sec^2 \theta) \, d\theta $$ Using the identity $ 1 + \tan^2 \theta = \sec^2 \theta $: $$ I = \int_{\pi/4}^{\pi/3} \frac{\tan \theta}{(\sec^2 \theta)^2} \cdot 2 \tan \theta \sec^2 \theta \, d\theta $$ $$ I = \int_{\pi/4}^{\pi/3} \frac{2 \tan^2 \theta \sec^2 \theta}{\sec^4 \theta} \, d\theta = 2 \int_{\pi/4}^{\pi/3} \frac{\tan^2 \theta}{\sec^2 \theta} \, d\theta $$ Convert to sine and cosine: $$ I = 2 \int_{\pi/4}^{\pi/3} \frac{\sin^2 \theta / \cos^2 \theta}{1 / \cos^2 \theta} \, d\theta = 2 \int_{\pi/4}^{\pi/3} \sin^2 \theta \, d\theta $$

Step 3: Integrating and applying limits. Use the power-reduction identity $ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} $: $$ I = 2 \int_{\pi/4}^{\pi/3} \frac{1 - \cos 2\theta}{2} \, d\theta = \int_{\pi/4}^{\pi/3} (1 - \cos 2\theta) \, d\theta $$ $$ I = \left[ \theta - \frac{\sin 2\theta}{2} \right]_{\pi/4}^{\pi/3} $$ Evaluate at the upper limit ($ \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} $.
Evaluate at the lower limit ($ \pi/4 $): $ \frac{\pi}{4} - \frac{\sin(2\pi/4)}{2} = \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} = \frac{\pi}{4} - \frac{1}{2} $.
Subtracting the two: $$ I = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right) $$ $$ I = \frac{\pi}{3} - \frac{\pi}{4} + \frac{1}{2} - \frac{\sqrt{3}}{4} = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} $$