Question #33600

Feb 28, 2017

$\frac{1}{2}$.

Explanation:

Let $I = {\int}_{1}^{9} \frac{1}{\sqrt{x} {\left(1 + \sqrt{x}\right)}^{2}} \mathrm{dx} .$

Let us subst. $\sqrt{x} = t , \mathmr{and} , x = {t}^{2} , \text{ so that, } \mathrm{dx} = 2 t \mathrm{dt} .$

Also, $x = 1 \Rightarrow t = \sqrt{x} = \sqrt{1} = 1 , \text{ &, for } x = 9 , t = 3.$

Hence, $I = {\int}_{1}^{3} \frac{2 t}{t {\left(1 + t\right)}^{2}} \mathrm{dt} = 2 {\int}_{1}^{3} {\left(1 + t\right)}^{-} 2 \mathrm{dt} .$

Now, recall the following Useful Result :

$\int f \left(x\right) \mathrm{dx} = F \left(x\right) + c \Rightarrow \int f \left(a x + b\right) \mathrm{dx} = \frac{1}{a} F \left(a x + b\right) + c ' , a \ne 0.$

Knowing that, $\int {t}^{-} 2 \mathrm{dt} = - \frac{1}{t} + C ' ,$ we have,

$I = 2 {\left[- \frac{1}{1 + t}\right]}_{1}^{3} = - 2 \left[\frac{1}{4} - \frac{1}{2}\right] = - 2 \left(- \frac{1}{4}\right) = \frac{1}{2.}$

Enjoy Maths.!

Feb 28, 2017

$\frac{1}{2}$

Explanation:

Making $y = \sqrt{x}$ and consequently $\mathrm{dy} = \frac{1}{2} \frac{\mathrm{dx}}{y}$ we have

$\int \frac{1}{\sqrt{x} \cdot {\left(1 + \sqrt{x}\right)}^{2}} \mathrm{dx} = \int \frac{2 y}{y {\left(1 + y\right)}^{2}} \mathrm{dy} = 2 \int \frac{\mathrm{dy}}{1 + y} ^ 2 = - \frac{2}{1 + y} + C$

but

${\int}_{1}^{9} \frac{1}{\sqrt{x} \cdot {\left(1 + \sqrt{x}\right)}^{2}} \mathrm{dx} = 2 {\int}_{1}^{\sqrt{9}} \frac{\mathrm{dy}}{1 + y} ^ 2 = \frac{1}{2}$