What is the integral from 1 to 2 of x^{-1}+x^4 ?

The integral from 1 to 2 of x^{-1}+x^4 is ln(2)+31/5

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integral from 1 to 2 of x^{-1}+x^4

\int_{1}^{2} x^{- 1} + x^{4} d x

ln (2) + \frac{31}{5}

Decimal

6.89314 \dots

Solution steps

\int_{1}^{2} x^{- 1} + x^{4} d x

Apply exponent rule:

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

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

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

\int_{1}^{2} x^{4} d x = \frac{31}{5}

= ln (2) + \frac{31}{5}

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What is the integral from 1 to 2 of x^{-1}+x^4 ?
The integral from 1 to 2 of x^{-1}+x^4 is ln(2)+31/5