Evaluate the integral of arctan x1/2

Evaluate the following integral

$\int \arctan \sqrt{x} \, dx.$

We use integration by parts, letting

$\begin{align*} u &= \arctan \sqrt{x} & du &= \frac{1}{2} \left( \frac{1}{\sqrt{x} (1+x)} \right) \, dx \\ dv &= dx & v &= x. \end{align*}$

Therefore,

$\int \arctan \sqrt{x} \, dx &= x \arctan \sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{1+x} \, dx.$

Now, to evaluate the integral $\int \frac{\sqrt{x}}{1+x} \, dx$ we make the substitution, $s = \sqrt{x}$ which implies $ds = \frac{1}{2\sqrt{x}} \, dx$ . Therefore,

$\begin{align*} \int \frac{\sqrt{x}}{1+x} \, dx &= 2 \int \frac{s^2}{1+s^2} \, ds \\ &= 2 \int \left(1 - \frac{1}{1+s^2}\right) \, ds \\ &= 2 (s - \arctan s) + C \\ &= 2(\sqrt{x} - \arctan \sqrt{x} ) + C. \end{align*}$

Plugging this back in to the equation above we obtain,

$\begin{align*} \int \arctan \sqrt{x} \, dx &= x \arctan \sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{1+x} \, dx \\[9pt] &= x \arctan \sqrt{x} - \frac{1}{2} \left( 2 (\sqrt{x} - \arctan \sqrt{x}) \right) + C \\[9pt] &= x \arctan \sqrt{x} - \sqrt{x} + \arctan \sqrt{x} + C \\[9pt] &= (x+1) \arctan \sqrt{x} - \sqrt{x} + C. \end{align*}$

Stumbling Robot

Evaluate the integral of arctan x^1/2

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