What is the integral from 1 to 3 of 1/(x^2)-6/(x^3) ?

The integral from 1 to 3 of 1/(x^2)-6/(x^3) is -2

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integral from 1 to 3 of 1/(x^2)-6/(x^3)

\int_{1}^{3} \frac{1}{x ^{2}} - \frac{6}{x ^{3}} d x

- 2

Solution steps

\int_{1}^{3} \frac{1}{x ^{2}} - \frac{6}{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}^{3} \frac{1}{x ^{2}} d x - \int_{1}^{3} \frac{6}{x ^{3}} d x

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

\int_{1}^{3} \frac{6}{x ^{3}} d x = \frac{8}{3}

= \frac{2}{3} - \frac{8}{3}

Simplify

\frac{2}{3} - \frac{8}{3} : - 2

= - 2

\frac{d}{d x} (\frac{10}{x ^{2}})

\frac{d}{d x} (sin (9 x^{2}))

n = 1 \sum \infty \frac{1}{n} cos (nπ)

f (x) = x^{2} + 2 x, x = 3, x = 5

f (x) = ln (x^{2} + 3)

What is the integral from 1 to 3 of 1/(x^2)-6/(x^3) ?
The integral from 1 to 3 of 1/(x^2)-6/(x^3) is -2