The value of the definite integral $I = \int_{1}^{2} \frac{1}{x(1 + x^2)}dx$ is:

The definite integral to evaluate is:

$I = \int_{1}^{2} \frac{1}{x(1 + x^2)}dx$

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Using partial fractions to decompose:

$\frac{1}{x(1 + x^2)} = \frac{A}{x} + \frac{Bx + C}{1 + x^2}$

Multiply both sides by $x(1 + x^2)$:

$1 = A(1 + x^2) + (Bx + C)x$

$1 = A + Ax^2 + Bx^2 + Cx$

$1 = A + (A + B)x^2 + Cx$

Matching coefficients:

Constant term: $A = 1$

Coefficient of $x$: $C = 0$

Coefficient of $x^2$: $A + B = 0$, so $B = -1$

Therefore:

$\frac{1}{x(1 + x^2)} = \frac{1}{x} - \frac{x}{1 + x^2}$

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The integral becomes:

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

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

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Evaluating at $x = 2$:

$\ln 2 - \frac{1}{2}\ln(1 + 4)$

$= \ln 2 - \frac{1}{2}\ln 5$

Evaluating at $x = 1$:

$\ln 1 - \frac{1}{2}\ln(1 + 1)$

$= 0 - \frac{1}{2}\ln 2$

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Computing the definite integral:

$I = \left(\ln 2 - \frac{1}{2}\ln 5\right) - \left(0 - \frac{1}{2}\ln 2\right)$

$I = \ln 2 - \frac{1}{2}\ln 5 + \frac{1}{2}\ln 2$

$I = \frac{3}{2}\ln 2 - \frac{1}{2}\ln 5$

Therefore: $I = \dfrac{3}{2}\log 2 - \dfrac{1}{2}\log 5$