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

Concept: Since $f(x)$ is defined as an indefinite integral, the quantity $f(3)-f(1)$ can be evaluated directly using the Fundamental Theorem of Calculus: \[ f(3)-f(1)=\int_{1}^{3}\frac{\sqrt{x}}{(1+x)^2}\,dx \] This integral can be solved conveniently using the substitution $x=\tan^2\theta$.

Step 1: Applying trigonometric substitution.
Let \[ \sqrt{x}=\tan\theta \] Then, \[ x=\tan^2\theta \] Differentiating: \[ dx=2\tan\theta\sec^2\theta\,d\theta \] Also, \[ 1+x=1+\tan^2\theta=\sec^2\theta \] Now change the limits: When $x=1$, \[ \tan\theta=1 \Rightarrow \theta=\frac{\pi}{4} \] When $x=3$, \[ \tan\theta=\sqrt3 \Rightarrow \theta=\frac{\pi}{3} \]

Step 2: Transforming the integral completely.
Substitute all expressions into the integral: \[ I=\int_{\pi/4}^{\pi/3} \frac{\tan\theta}{(\sec^2\theta)^2} \cdot 2\tan\theta\sec^2\theta\,d\theta \] Simplifying carefully: \[ I= \int_{\pi/4}^{\pi/3} \frac{2\tan^2\theta\sec^2\theta}{\sec^4\theta}\,d\theta \] \[ I= \int_{\pi/4}^{\pi/3} 2\tan^2\theta\cos^2\theta\,d\theta \] Using \[ \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta} \] we get \[ I= \int_{\pi/4}^{\pi/3} 2\sin^2\theta\,d\theta \] Now use the identity \[ 2\sin^2\theta=1-\cos2\theta \] Hence, \[ I= \int_{\pi/4}^{\pi/3} (1-\cos2\theta)\,d\theta \]

Step 3: Integrating and evaluating the limits.
Integrating: \[ I= \left[ \theta-\frac{\sin2\theta}{2} \right]_{\pi/4}^{\pi/3} \] Substituting upper and lower limits: \[ I= \left( \frac{\pi}{3}-\frac{\sin(2\pi/3)}{2} \right) - \left( \frac{\pi}{4}-\frac{\sin(\pi/2)}{2} \right) \] Using standard trigonometric values: \[ \sin\frac{2\pi}{3}=\frac{\sqrt3}{2}, \qquad \sin\frac{\pi}{2}=1 \] Therefore, \[ I= \left( \frac{\pi}{3}-\frac{\sqrt3}{4} \right) - \left( \frac{\pi}{4}-\frac12 \right) \] \[ I= \frac{\pi}{3}-\frac{\pi}{4} +\frac12 -\frac{\sqrt3}{4} \] \[ I= \frac{\pi}{12} +\frac12 -\frac{\sqrt3}{4} \] Hence, \[ \boxed{ f(3)-f(1) = \frac{\pi}{12} +\frac12 -\frac{\sqrt3}{4} } \]