# How do you find the antiderivative of int (arctan(sqrtx)) dx?

Nov 7, 2016

$\left(x + 1\right) \arctan \left(\sqrt{x}\right) - \sqrt{x} + C$

#### Explanation:

$I = \int \arctan \left(\sqrt{x}\right) \mathrm{dx}$

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. So, for $\int \arctan \left(\sqrt{x}\right) \mathrm{dx}$, we should let $u = \arctan \left(\sqrt{x}\right)$ and $\mathrm{dv} = \mathrm{dx}$.

Differentiating $u$ gives $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {\left(\sqrt{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \sqrt{x}$ or $\mathrm{du} = \frac{1}{2 \sqrt{x} \left(1 + x\right)} \mathrm{dx}$. From $\mathrm{dv} = \mathrm{dx}$ we integrate to show that $v = x$. Thus:

$I = u v - \int v \mathrm{du}$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \int \frac{x}{2 \sqrt{x} \left(1 + x\right)} \mathrm{dx}$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} \mathrm{dx}$

Letting $t = \sqrt{x}$ implies that ${t}^{2} = x$ and furthermore, that $2 t \mathrm{dt} = \mathrm{dx}$ through differentiating. Thus:

$I = x \arctan \left(\sqrt{x}\right) - \frac{1}{2} \int \frac{t}{1 + {t}^{2}} \left(2 t \mathrm{dt}\right)$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \int {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt}$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \int \frac{1 + {t}^{2} - 1}{1 + {t}^{2}} \mathrm{dt}$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \int \frac{1 + {t}^{2}}{1 + {t}^{2}} \mathrm{dt} + \int \frac{1}{1 + {t}^{2}} \mathrm{dt}$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - \int \mathrm{dt} + \arctan \left(t\right)$

$\textcolor{w h i t e}{I} = x \arctan \left(\sqrt{x}\right) - t + \arctan \left(t\right)$

Since $t = \sqrt{x}$:

$I = x \arctan \left(\sqrt{x}\right) - \sqrt{x} + \arctan \left(\sqrt{x}\right)$

$\textcolor{w h i t e}{I} = \left(x + 1\right) \arctan \left(\sqrt{x}\right) - \sqrt{x} + C$