What is the integral from 1 to e of x^5+x^{-5}+1/x ?

The integral from 1 to e of x^5+x^{-5}+1/x is 5/4+(2e^{10}-2e^4-3)/(12e^4)

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integral from 1 to e of x^5+x^{-5}+1/x

\int_{1}^{e} x^{5} + x^{- 5} + \frac{1}{x} d x

\frac{5}{4} + \frac{2 e ^{10} - 2 e ^{4} - 3}{12 e ^{4}}

Decimal

68.31688 \dots

Solution steps

\int_{1}^{e} x^{5} + x^{- 5} + \frac{1}{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}^{e} x^{5} d x + \int_{1}^{e} x^{- 5} d x + \int_{1}^{e} \frac{1}{x} d x

\int_{1}^{e} x^{5} d x = \frac{e ^{6} - 1}{6}

\int_{1}^{e} x^{- 5} d x = - \frac{1}{4 e ^{4}} + \frac{1}{4}

\int_{1}^{e} \frac{1}{x} d x = 1

= \frac{e ^{6} - 1}{6} - \frac{1}{4 e ^{4}} + \frac{1}{4} + 1

Combine the fractions

1 + \frac{1}{4} : \frac{5}{4}

Combine the fractions

\frac{e ^{6} - 1}{6} - \frac{1}{4 e ^{4}} : \frac{2 e ^{10} - 2 e ^{4} - 3}{12 e ^{4}}

= \frac{5}{4} + \frac{2 e ^{10} - 2 e ^{4} - 3}{12 e ^{4}}

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What is the integral from 1 to e of x^5+x^{-5}+1/x ?
The integral from 1 to e of x^5+x^{-5}+1/x is 5/4+(2e^{10}-2e^4-3)/(12e^4)