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

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

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

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

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

Decimal

1.19314 \dots

Solution steps

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

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

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

= ln (2) + \frac{1}{2}

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