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

The integral from 1 to 2 of [ 1/x-2/(x^3)] is ln(2)-3/4

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integral from 1 to 2 of [ 1/x-2/(x^3)]

\int_{1}^{2} [\frac{1}{x} - \frac{2}{x ^{3}}] d x

ln (2) - \frac{3}{4}

Decimal

- 0.05685 \dots

Solution steps

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

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

\int_{1}^{2} \frac{2}{x ^{3}} d x = \frac{3}{4}

= ln (2) - \frac{3}{4}

\int_{0}^{\frac{π}{2}} 0 d x

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

\int_{0}^{1} x - x^{5} d x

\int_{- 1}^{1} t^{5} d t

\int_{0}^{2} (8 - 2 x^{2}) d x

What is the integral from 1 to 2 of [ 1/x-2/(x^3)] ?
The integral from 1 to 2 of [ 1/x-2/(x^3)] is ln(2)-3/4