# How do you evaluate the integral int arctansqrtx?

##### 1 Answer
Jan 15, 2017

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

#### Explanation:

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

Let $t = \sqrt{x}$. Finding $\mathrm{dx}$ in a usable form is simpler if we first write that ${t}^{2} = x$, which then implies that $2 t \mathrm{dt} = \mathrm{dx}$. This way we can plug in $\mathrm{dx}$ into the integral straight away. These substitutions yield:

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

Now use integration by parts. Let:

$\left\{\begin{matrix}u = \arctan \left(t\right) \text{ "=>" "du=1/(t^2+1)dt \\ dv=2tdt" "=>" } v = {t}^{2}\end{matrix}\right.$

Then:

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

Rewriting the integrand as $\frac{{t}^{2} + 1 - 1}{{t}^{2} + 1} = \frac{{t}^{2} + 1}{{t}^{2} + 1} - \frac{1}{{t}^{2} + 1} = 1 - \frac{1}{{t}^{2} + 1}$ we see that

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

Both of which are common integrals:

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

Returning to $x$ from $t = \sqrt{x}$:

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

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