$\int_{\pi/6}^{\pi/3} \frac{tan x}{tan x + cot x} dx$ is equal to

The integrand can be simplified:

$\frac{\tan x}{\tan x + \cot x}$

Since $\cot x = \frac{1}{\tan x}$:

$\frac{\tan x}{\tan x + \cot x} = \frac{\tan x}{\tan x + \frac{1}{\tan x}}$

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Multiply both numerator and denominator by $\tan x$:

$\frac{\tan x \cdot \tan x}{\left(\tan x + \frac{1}{\tan x}\right) \cdot \tan x}$

$= \frac{\tan^2 x}{\tan^2 x + 1}$

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Using the identity $\tan^2 x + 1 = \sec^2 x$:

$\frac{\tan^2 x}{\tan^2 x + 1} = \frac{\tan^2 x}{\sec^2 x}$

Since $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$:

$\frac{\tan^2 x}{\sec^2 x} = \frac{\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}}$

$= \sin^2 x$

The integral becomes:

$\int_{\pi/6}^{\pi/3} \sin^2 x \, dx$

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Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:

$\int_{\pi/6}^{\pi/3} \sin^2 x \, dx = \int_{\pi/6}^{\pi/3} \frac{1 - \cos 2x}{2} \, dx$

$= \frac{1}{2} \int_{\pi/6}^{\pi/3} (1 - \cos 2x) \, dx$

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Integrating:

$= \frac{1}{2} \left[x - \frac{\sin 2x}{2}\right]_{\pi/6}^{\pi/3}$

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Evaluating at the limits:

At $x = \pi/3$: $\sin(2\pi/3) = \frac{\sqrt{3}}{2}$

At $x = \pi/6$: $\sin(\pi/3) = \frac{\sqrt{3}}{2}$

$= \frac{1}{2} \left[\left(\frac{\pi}{3} - \frac{\sqrt{3}/2}{2}\right) - \left(\frac{\pi}{6} - \frac{\sqrt{3}/2}{2}\right)\right]$

The $\frac{\sqrt{3}/2}{2}$ terms cancel:

$= \frac{1}{2} \left[\frac{\pi}{3} - \frac{\pi}{6}\right]$

$= \frac{1}{2} \left[\frac{2\pi - \pi}{6}\right]$

$= \frac{1}{2} \cdot \frac{\pi}{6}$

$= \frac{\pi}{12}$

Therefore, the answer is $\frac{\pi}{12}$.