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

The integral from 1 to 3 of (6-2x)/(x^3) is 4/3

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

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

\frac{4}{3}

Solution steps

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

Expand

\frac{6 - 2 x}{x ^{3}} : \frac{6}{x ^{3}} - \frac{2}{x ^{2}}

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

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

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

= \frac{8}{3} - \frac{4}{3}

Simplify

\frac{8}{3} - \frac{4}{3} : \frac{4}{3}

= \frac{4}{3}

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What is the integral from 1 to 3 of (6-2x)/(x^3) ?
The integral from 1 to 3 of (6-2x)/(x^3) is 4/3