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

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

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

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

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

Decimal Notation

14.39953 \dots

Solution steps

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

Expand

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

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

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} t^{3} d t + \int_{1}^{e} \frac{1}{t} d t

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

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

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

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