What is the integral from 1 to e of (x^4+1)/x ?

The integral from 1 to e of (x^4+1)/x is (e^4-1)/4+1

integral from 1 to e of (x^4+1)/x

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

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

Decimal

14.39953 \dots

Solution steps

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

Expand

\frac{x ^{4} + 1}{x} : x^{3} + \frac{1}{x}

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

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

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

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

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What is the integral from 1 to e of (x^4+1)/x ?
The integral from 1 to e of (x^4+1)/x is (e^4-1)/4+1