# Question #362bd

Nov 18, 2016

$2 {\log}_{e} \left(\sqrt{x} + 1\right)$

#### Explanation:

Note that $\frac{1}{x + \sqrt{x}} = \frac{1}{\sqrt{x} \left(\sqrt{x} + 1\right)}$ and also that
$\frac{d}{\mathrm{dx}} \left(\sqrt{x} + 1\right) = \frac{1}{2} \left(\frac{1}{\sqrt{x}}\right)$ so

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

$\int \frac{\mathrm{dx}}{x + \sqrt{x}} = \int \frac{1}{x + \sqrt{x}} \mathrm{dx} = \int 2 \frac{\frac{d}{\mathrm{dx}} \left(\sqrt{x} + 1\right)}{\sqrt{x} + 1} \mathrm{dx} = 2 {\log}_{e} \left(\sqrt{x} + 1\right)$

Nov 18, 2016

The answer is $= 2 \ln \left(1 + \sqrt{x}\right) + C$

#### Explanation:

The two writing represent the same integral.

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

Multiply numerator and denominator by $x - \sqrt{x}$

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

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

Solving by substitution

Let $u = \sqrt{x}$ ;
${u}^{2} = x$

${u}^{4} = {x}^{2}$

$\mathrm{du} = \frac{\mathrm{dx}}{2 \sqrt{x}}$

$\mathrm{dx} = 2 u \mathrm{du}$

$I = \int \left({u}^{2} - u\right) \frac{2 u \mathrm{du}}{{u}^{4} - {u}^{2}}$

$= 2 \int \frac{{u}^{2} \left(u - 1\right) \mathrm{du}}{{u}^{2} \left({u}^{2} - 1\right)}$

$= 2 \int \frac{\cancel{u - 1} \mathrm{du}}{\left(u + 1\right) \cancel{u - 1}}$

$= 2 \ln \left(u + 1\right)$

$= 2 \ln \left(1 + \sqrt{x}\right) + C$