let f(x)
Question:

Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x \geq 0)$. Then $f(3)-f(1)$ is equal to :

  1. (1) $-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

  2. (2) $\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

  3. (3) $-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

  4. (4) $\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$


Correct Option: , 4

Solution:

$\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x>0)$

Put $x=\tan ^{2} \theta \Rightarrow 2 x d x=2 \tan \theta \sec ^{2} \theta d \theta$

$I=\int \frac{2 \tan ^{2} \theta \cdot \sec ^{2} \theta}{\sec ^{4} \theta} d \theta=\int 2 \sin ^{2} \theta d \theta$

$=\theta-\frac{\sin 2 \theta}{2}+C$

$\Rightarrow f(x)=\theta-\frac{1}{2} \times \frac{2 \tan \theta}{1+\tan ^{2} \theta}+C$

$f(x)=\theta-\frac{\tan \theta}{1+\tan ^{2} \theta}+C=\tan ^{-1} \sqrt{x}-\frac{\sqrt{x}}{1+x}+C$

Now $f(3)-f(1)=\tan ^{-1}(\sqrt{3})-\frac{\sqrt{3}}{1+3}-\tan ^{-1}(1)+\frac{1}{2}$

$=\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.