# How do you find the integral int (sqrtx-1)^2/sqrtxdx using substitution?

Aug 24, 2016

$\int {\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} \mathrm{dx} = \frac{2}{3} {\left(\sqrt{x} - 1\right)}^{3} + C$

#### Explanation:

Calling $y = \sqrt{x}$ after

$\mathrm{dy} = \frac{1}{2} \frac{\mathrm{dx}}{\sqrt{x}}$

and substituting

$\int {\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} \mathrm{dx} \equiv 2 \int {\left(y - 1\right)}^{2} \mathrm{dy} = \frac{2}{3} {\left(y - 1\right)}^{3} + C$

Recovering the initial variable

$\int {\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} \mathrm{dx} = \frac{2}{3} {\left(\sqrt{x} - 1\right)}^{3} + C$

Aug 24, 2016

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

#### Explanation:

Let $I = \int {\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} \mathrm{dx}$

We subst. $x = {t}^{2} \Rightarrow \mathrm{dx} = 2 t \mathrm{dt}$. Also,

${\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} = {\left(t - 1\right)}^{2} / t = \frac{{t}^{2} - 2 t + 1}{t}$.

$\Rightarrow I = \int \left\{\frac{{t}^{2} - 2 t + 1}{t}\right\} 2 t \mathrm{dt}$

$= 2 \int \left({t}^{2} - 2 t + 1\right) \mathrm{dt}$

$= 2 \left[{t}^{3} / 3 - 2 {t}^{2} / 2 + t\right]$

$= \frac{2 t}{3} \left({t}^{2} - 3 t + 3\right)$

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

Though we have solved the Problem using Substn. Methodas was

so demanded , but, in fact, there is no such need to solve it

in this fashion. Have a look :

$I = \int {\left(\sqrt{x} - 1\right)}^{2} / \sqrt{x} \mathrm{dx}$

$= \int \frac{\left(x - 2 \sqrt{x} + 1\right)}{\sqrt{x}} \mathrm{dx}$

$= \int \left\{\frac{x}{\sqrt{x}} - 2 \frac{\sqrt{x}}{\sqrt{x}} + \frac{1}{\sqrt{x}}\right\} \mathrm{dx}$

$= \int \left({x}^{\frac{1}{2}} - 2 + {x}^{- \frac{1}{2}}\right) \mathrm{dx}$

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

$= \frac{2 {x}^{\frac{3}{2}}}{3} - 2 x + 2 {x}^{\frac{1}{2}}$

$= \frac{2 {x}^{\frac{1}{2}}}{3} \left(x - 3 {x}^{\frac{1}{2}} + 3\right) , \mathmr{and} , \frac{2 \sqrt{x}}{3} \left(x - 3 \sqrt{x} + 3\right) + C$, as before!

Enjoy Maths.!