What is the integral from 1 to 2 of (1+3x^2)/x ?

The integral from 1 to 2 of (1+3x^2)/x is ln(2)+9/2

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integral from 1 to 2 of (1+3x^2)/x

\int_{1}^{2} \frac{1 + 3 x ^{2}}{x} d x

ln (2) + \frac{9}{2}

Decimal

5.19314 \dots

Solution steps

\int_{1}^{2} \frac{1 + 3 x ^{2}}{x} d x

Expand

\frac{1 + 3 x ^{2}}{x} : \frac{1}{x} + 3 x

= \int_{1}^{2} \frac{1}{x} + 3 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{1}{x} d x + \int_{1}^{2} 3 x d x

\int_{1}^{2} \frac{1}{x} d x = ln (2)

\int_{1}^{2} 3 x d x = \frac{9}{2}

= ln (2) + \frac{9}{2}

What is the integral from 1 to 2 of (1+3x^2)/x ?
The integral from 1 to 2 of (1+3x^2)/x is ln(2)+9/2