Question #07873

Mar 4, 2016

$\int \frac{1}{x + \sqrt{x}} \mathrm{dx} = 2 \ln \left(\sqrt{x} + 1\right) + C$, with $x \ge 0$ and where $C$ is the constant of integration.

Explanation:

We want to compute the antiderivative of $\int \frac{1}{x + \sqrt{x}} \mathrm{dx}$.
Since we have $\sqrt{x}$, we'll assume $x \ge 0$.

We will use this change of variable :
$x = \phi \left(t\right) = {t}^{2}$

$\int \frac{1}{x + \sqrt{x}} \mathrm{dx}$ = $\int \frac{1}{{t}^{2} + \sqrt{{t}^{2}}} \cdot \dot{\left({t}^{2}\right)} \mathrm{dt}$

$= \int \frac{2 t}{{t}^{2} + t} \mathrm{dt} = 2 \int \frac{1}{t + 1} \mathrm{dt} = 2 \ln \left(| t + 1 |\right) + C .$

Since $x = {t}^{2}$,$t = \sqrt{x}$ for $x \ge 0$ (which was supposed above).

So $\int \frac{1}{x + \sqrt{x}} \mathrm{dx} = 2 \ln \left(| \sqrt{x} + 1 |\right) + C = 2 \ln \left(\sqrt{x} + 1\right) + C$, where $C$ is the constant of integration.