$\int_{1}^{2} \frac{1}{x(x+1)} dx, x > 0$ equals

To split $\frac{1}{x(x+1)}$ into simpler parts:

$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$

Multiply both sides by $x(x+1)$:

$1 = A(x+1) + Bx$

Put $x = 0$:

$1 = A(0+1) + B(0)$

$1 = A$

Put $x = -1$:

$1 = A(-1+1) + B(-1)$

$1 = -B$

$B = -1$

Therefore:

$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$

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$\int_{1}^{2} \frac{1}{x(x+1)} dx = \int_{1}^{2} \left(\frac{1}{x} - \frac{1}{x+1}\right) dx$

$= \left[\ln x - \ln(x+1)\right]_{1}^{2}$

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At $x = 2$: $\ln(2) - \ln(3)$

At $x = 1$: $\ln(1) - \ln(2)$

$= [\ln(2) - \ln(3)] - [\ln(1) - \ln(2)]$

$= \ln(2) - \ln(3) - \ln(1) + \ln(2)$

$= 2\ln(2) - \ln(3)$

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Using logarithm properties:

$= \ln(2^2) - \ln(3)$

$= \ln(4) - \ln(3)$

$= \ln\left(\frac{4}{3}\right)$

Therefore, the answer is $\ln\left(\frac{4}{3}\right)$.