What is the integral from 1 to 3 of x^{-1}+x^2 ?

The integral from 1 to 3 of x^{-1}+x^2 is ln(3)+26/3

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integral from 1 to 3 of x^{-1}+x^2

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

ln (3) + \frac{26}{3}

Decimal

9.76527 \dots

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

\int_{1}^{3} \frac{1}{x} d x = ln (3)

\int_{1}^{3} x^{2} d x = \frac{26}{3}

= ln (3) + \frac{26}{3}

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What is the integral from 1 to 3 of x^{-1}+x^2 ?
The integral from 1 to 3 of x^{-1}+x^2 is ln(3)+26/3