For $x \in \left(0, \frac{\pi}{2}\right)$, $\int \frac{1}{\sin^2 x + \sin 2x} dx$ is equal to

The denominator is $\sin^2 x + \sin 2x$.

Using $\sin 2x = 2\sin x \cos x$:

$\sin^2 x + 2\sin x \cos x$

$= \sin x(\sin x + 2\cos x)$

The integral becomes:

$\int \frac{1}{\sin x(\sin x + 2\cos x)} dx$

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Dividing both numerator and denominator by $\sin^2 x$:

$\int \frac{1/\sin^2 x}{\sin x(\sin x + 2\cos x)/\sin^2 x} dx$

The numerator becomes:

$\frac{1}{\sin^2 x}$

$= \csc^2 x$

The denominator becomes:

$\frac{\sin x(\sin x + 2\cos x)}{\sin^2 x}$

$= \frac{\sin x + 2\cos x}{\sin x}$

$= 1 + 2\cdot\frac{\cos x}{\sin x}$

$= 1 + 2\cot x$

The integral is now:

$\int \frac{\csc^2 x}{1 + 2\cot x} dx$

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Let $u = \cot x$

Then $\frac{du}{dx} = -\csc^2 x$, so $du = -\csc^2 x \, dx$

Therefore $\csc^2 x \, dx = -du$

Substituting:

$\int \frac{\csc^2 x}{1 + 2\cot x} dx$

$= \int \frac{-du}{1 + 2u}$

$= -\int \frac{du}{1 + 2u}$

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$-\int \frac{du}{1 + 2u}$

$= -\frac{1}{2}\ln|1 + 2u| + C$

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Substituting back $u = \cot x = \frac{1}{\tan x}$:

$-\frac{1}{2}\ln\left|1 + \frac{2}{\tan x}\right| + C$

$= -\frac{1}{2}\ln\left|\frac{\tan x + 2}{\tan x}\right| + C$

Using the logarithm property $-\ln(a) = \ln\left(\frac{1}{a}\right)$:

$= \frac{1}{2}\ln\left|\frac{\tan x}{\tan x + 2}\right| + C$

Therefore:

$\boxed{\frac{1}{2}\log\left|\frac{\tan x}{\tan x + 2}\right| + C}$