What is the integral from 1 to 5 of (2/(x^2)+x) ?

The integral from 1 to 5 of (2/(x^2)+x) is 68/5

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integral from 1 to 5 of (2/(x^2)+x)

\int_{1}^{5} (\frac{2}{x ^{2}} + x) d x

\frac{68}{5}

Solution steps

\int_{1}^{5} \frac{2}{x ^{2}} + x 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}^{5} \frac{2}{x ^{2}} d x + \int_{1}^{5} x d x

\int_{1}^{5} \frac{2}{x ^{2}} d x = \frac{8}{5}

\int_{1}^{5} x d x = 12

= \frac{8}{5} + 12

Simplify

\frac{8}{5} + 12 : \frac{68}{5}

= \frac{68}{5}

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What is the integral from 1 to 5 of (2/(x^2)+x) ?
The integral from 1 to 5 of (2/(x^2)+x) is 68/5