# Question #31e26

Dec 4, 2016

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

#### Explanation:

It is difficult to work with the $\sqrt{x}$ as the argument of $\arctan$, so we begin by making a substitution.

Let $t = \sqrt{x} \implies \mathrm{dt} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$. Then

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

$= \int 2 t \arctan \left(t\right) \mathrm{dt}$

Next, we will apply integration by parts. To apply the formula $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, we let

$u = \arctan \left(t\right)$ and $\mathrm{dv} = 2 t \mathrm{dt}$
$\implies \mathrm{du} = \frac{1}{1 + {t}^{2}} \mathrm{dt}$ and $v = {t}^{2}$

Applying the formula, this gives

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

Focusing on the remaining integral, we have

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

$= \int \mathrm{dt} - \int \frac{1}{1 + {t}^{2}} \mathrm{dt}$

$= t - \arctan \left(t\right) + C$

Putting this all together, we get our final result:

$\int \arctan \left(\sqrt{x}\right) \mathrm{dx} = \int 2 t \arctan \left(t\right) \mathrm{dt}$

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

$= {t}^{2} \arctan \left(t\right) - t + \arctan \left(t\right) + C$

$= \left({t}^{2} + 1\right) \arctan \left(t\right) - t + C$

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