# Question #4d43b

Aug 17, 2016

$\frac{2}{3} \left\{{\left(1 + x\right)}^{\frac{3}{2}} - {x}^{\frac{3}{2}}\right\} + C$.

#### Explanation:

Let $I = \int \frac{1}{\sqrt{1 + x} + \sqrt{x}} \mathrm{dx}$

$\therefore I = \int \frac{1}{\sqrt{1 + x} + \sqrt{x}} \times \frac{\sqrt{1 + x} - \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}} \mathrm{dx}$

$= \int \frac{\sqrt{1 + x} - \sqrt{x}}{1 + x - x} \mathrm{dx} = \int \left(\sqrt{1 + x} - \sqrt{x}\right) \mathrm{dx}$

$= \int \sqrt{1 + x} \mathrm{dx} - \int \sqrt{x} \mathrm{dx} = {I}_{1} - \int {x}^{\frac{1}{2}} \mathrm{dx}$

$= {I}_{1} - {x}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) = {I}_{1} - \frac{2}{3} {x}^{\frac{3}{2}}$, where,

${I}_{1} = \int \sqrt{1 + x} \mathrm{dx}$

Let, $1 + x = {t}^{2} , s o , \mathrm{dx} = 2 t \mathrm{dt}$. Hence,

${I}_{1} = \int \sqrt{{t}^{2}} \cdot 2 t \mathrm{dt}$

$= 2 \int {t}^{2} \mathrm{dt} = 2 \cdot {t}^{3} / 3 = \frac{2}{3} {\left({t}^{2}\right)}^{\frac{3}{2}} = \frac{2}{3} {\left(1 + x\right)}^{\frac{3}{2}}$.

Altogether,

$I = {I}_{1} - \frac{2}{3} {x}^{\frac{3}{2}} = \frac{2}{3} {\left(1 + x\right)}^{\frac{3}{2}} - \frac{2}{3} {x}^{\frac{3}{2}}$, i.e.,

$I = \frac{2}{3} \left\{{\left(1 + x\right)}^{\frac{3}{2}} - {x}^{\frac{3}{2}}\right\} + C$.

In fact, we have used the substn. method because it was so

required. Otherwise, we can solve it without using substn., by the

following Rule :-

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

Now, $\int \sqrt{x} \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} + c \Rightarrow \int \sqrt{1 + x} = \frac{1}{1} \cdot \frac{2}{3} \cdot {\left(1 + x\right)}^{\frac{3}{2}} + k$.

Enjoy Maths.!