Question

Let \[ f(x)=\int \frac{\sqrt{x}}{(1+x)^2}\,dx, \quad (x \ge 0) \]

Then, find the value of: \[ f(3)-f(1) \] 

• 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:

Whenever an integral contains expressions like $\sqrt{x}$ together with $(1+x)$, the substitution \[ x=\tan^2\theta \] is extremely effective because it converts \[ 1+\tan^2\theta \] directly into \[ \sec^2\theta. \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks for the difference of an indefinite integral evaluated at two points, which is essentially a definite integral: \(\int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} dx\).
The integrand consists of a square root and a polynomial in the denominator. To solve this, we should use a trigonometric substitution to eliminate the radical and simplify the expression into something integrable.
Step 2: Key Formula or Approach:
The presence of \(1+x\) in the denominator suggests using \(x = \tan^2\theta\), because \(1 + \tan^2\theta = \sec^2\theta\).
Let \(x = \tan^2\theta \implies \sqrt{x} = \tan\theta\).
Differentiating both sides: \(dx = 2\tan\theta \sec^2\theta d\theta\).
Step 3: Detailed Explanation:
Substitute these into the integral:
\[ \int \frac{\tan\theta}{(\sec^2\theta)^2} (2\tan\theta \sec^2\theta) d\theta \]
\[ = \int \frac{2\tan^2\theta \sec^2\theta}{\sec^4\theta} d\theta = 2 \int \frac{\tan^2\theta}{\sec^2\theta} d\theta \]
Convert to sine and cosine:
\[ = 2 \int \frac{\sin^2\theta/\cos^2\theta}{1/\cos^2\theta} d\theta = 2 \int \sin^2\theta d\theta \]
Use the trigonometric identity \(2\sin^2\theta = 1 - \cos 2\theta\):
\[ = \int (1 - \cos 2\theta) d\theta = \theta - \frac{\sin 2\theta}{2} \]
Now, convert \(\sin 2\theta\) back to terms of \(\tan\theta\):
\(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\sqrt{x}}{1+x}\).
So, \(f(x) = \tan^{-1}\sqrt{x} - \frac{\sqrt{x}}{1+x}\).
Now calculate \(f(3) - f(1)\):
At \(x = 3\): \(f(3) = \tan^{-1}\sqrt{3} - \frac{\sqrt{3}}{1+3} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}\).
At \(x = 1\): \(f(1) = \tan^{-1}\sqrt{1} - \frac{\sqrt{1}}{1+1} = \frac{\pi}{4} - \frac{1}{2}\).
\[ f(3) - f(1) = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right) \]
\[ = \frac{\pi}{3} - \frac{\pi}{4} + \frac{1}{2} - \frac{\sqrt{3}}{4} \]
\[ = \frac{4\pi - 3\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} \]
Step 4: Final Answer:
The value is \(\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}\).
Hence, the correct option is (B).