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 Notation

0.19314 \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}

\int \frac{y}{y ^{2} + 2 y + 1} d y

x \to 0 lim (6 x - 1)

\int \frac{- 3}{x ^{2} x ^{2} - 9} d x

\int cos (ln (7 x)) d x

x \to 0 lim (- x^{6})

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