# How do I integrate intdx/(sqrt(x)(sqrt(x)+1))?

Oct 29, 2017

$\int \frac{\mathrm{dx}}{\sqrt{x} \cdot \left(\sqrt{x} + 1\right)} = 2 L n \left(\sqrt{x} + 1\right) + C$

#### Explanation:

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

After using $x = {\left(\tan u\right)}^{4}$ and $\mathrm{dx} = 4 {\left(\tan u\right)}^{3} \cdot {\left(\sec u\right)}^{2} \cdot \mathrm{du}$ transforms, $I$ became,

$I = \int \frac{4 {\left(\tan u\right)}^{3} \cdot {\left(\sec u\right)}^{2} \cdot \mathrm{du}}{{\left(\tan u\right)}^{2} \cdot {\left(\sec u\right)}^{2}}$

=$\int 4 \tan u \cdot \mathrm{du}$

=$4 \ln \left(\sec u\right) + C$

=$2 L n \left[{\left(\sec u\right)}^{2}\right] + C$

=$2 L n \left[{\left(\tan u\right)}^{2} + 1\right] + C$

After using $x = {\left(\tan u\right)}^{4}$ and ${\left(\tan u\right)}^{2} = \sqrt{x}$ inverse transforms, I found,

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

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

Note: This integral also can be found by $\sqrt{x} = u$, $x = {u}^{2}$ and $\mathrm{dx} = 2 u \cdot \mathrm{du}$ transforms,

Oct 29, 2017

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

#### Explanation:

Use the substitutions:

$u = \sqrt{x} + 1$

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

$\int \frac{\mathrm{dx}}{\sqrt{x} \left(\sqrt{x} + 1\right)} = 2 \int \frac{\mathrm{dx}}{2 \sqrt{x} \left(\sqrt{x} + 1\right)} = 2 \int \frac{\mathrm{du}}{u} = 2 \ln | u | + C$

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