Question

Hii guys solve the above attachment​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \: f(x) = \int \dfrac{ \sqrt{x} }{ \left( 1 + x\right)^{2} } \: dx$

$\bf{Put \: \: \: x = \tan^{2} ( \theta) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

$\bf{ \mapsto\: \: \: dx = 2 \tan( \theta) \cdot \sec^{2} ( \theta) \: d \theta }$

So,

$\displaystyle \implies\: f(x) = \int \dfrac{ \tan( \theta) }{ \left( 1 + \tan^{2} ( \theta) \right)^{2} } \cdot \: 2 \tan( \theta) \cdot \sec^{2} ( \theta) \: d \theta$

$\displaystyle \implies\: f(x) = 2 \int \dfrac{ \tan( \theta) }{ \sec^{4} ( \theta) } \cdot \tan( \theta) \cdot \sec^{2} ( \theta) \: d \theta$

$\displaystyle \implies\: f(x) = 2 \int \dfrac{ \tan^{2} ( \theta) }{ \sec^{2} ( \theta) } \: d \theta$

$\displaystyle \implies\: f(x) = 2 \int \sin^{2} ( \theta) \: d \theta$

$\displaystyle \implies\: f(x) = \int2 \sin^{2} ( \theta) \: d \theta$

$\displaystyle \implies\: f(x) = \int(1 - \cos( 2\theta) ) \: d \theta$

$\displaystyle \implies\: f(x) = \int \: d \theta- \int\cos( 2\theta) \: d \theta$

$\displaystyle \implies\: f(x) = \theta- \dfrac{\cos( 2\theta)}{2} + c$

From (1), we get,

$\displaystyle \implies\: f(x) = \tan^{ - 1} \left( \sqrt{x} \right) - \dfrac{1}{2} \cdot \dfrac{1 - x}{1 + x} + c$

Now,

$\displaystyle \implies\: f(3) = \tan^{ - 1} \left( \sqrt{3} \right) - \dfrac{1}{2} \cdot \dfrac{1 - 3}{1 + 3} + c$

$\displaystyle \implies\: f(3) = \dfrac{\pi}{3} + \dfrac{1 }{4} + c$

$\displaystyle \implies\: f(1) = \tan^{ - 1} \left( \sqrt{1} \right) - \dfrac{1}{2} \cdot \dfrac{1 - 1}{1 + x} + c$

$\displaystyle \implies\: f(1) = \dfrac{\pi}{4} + c$

So,

$\displaystyle \implies\: f(3) - f(1)= \dfrac{\pi}{3} + \dfrac{1 }{4} - \dfrac{\pi}{4}$

Answer 2

Step-by-step explanation:

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

$\\ \\ \text{Put \( \tt x=\tan ^{2} \theta \qquad \Rightarrow 2 x \: \: d x=2 \tan \theta \sec ^{2} \theta \: \: d \theta \)}$

$\\ \[ \begin{array}{l} \displaystyle \tt I=\int \frac{2 \tan ^{2} \theta \cdot \sec ^{2} \theta}{\sec ^{4} \theta} d \theta=\int 2 \sin ^{2} \theta d \theta \\ \\ \\ \\ \displaystyle \tt \quad=\theta-\frac{\sin 2 \theta}{2}+C \\ \\ \\ \\ \displaystyle \tt\Rightarrow f(x)=\theta-\frac{1}{2} \times \frac{2 \tan \theta}{1+\tan ^{2} \theta}+C \\ \\ \\ \\ \displaystyle \tt f(x)=\theta-\frac{\tan \theta}{1+\tan ^{2} \theta}+C=\tan ^{-1} \sqrt{x}-\frac{\sqrt{x}}{1+x}+C \end{array} \]$

$\\ \\ \[ \begin{aligned} \tt\text { Now } f(3)-f(1) &=\tan ^{-1}(\sqrt{3})-\frac{\sqrt{3}}{1+3}-\tan ^{-1}(1)+\frac{1}{2} \\ \\ \\ \\ &=\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4} \end{aligned} \]$