What is the integral from 1 to e of ((t^6+1))/t ?

The integral from 1 to e of ((t^6+1))/t is (e^6-1)/6+1

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integral from 1 to e of ((t^6+1))/t

\int_{1}^{e} \frac{( t ^{6} + 1 )}{t} d t

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

Decimal

68.07146 \dots

Solution steps

\int_{1}^{e} \frac{t ^{6} + 1}{t} d t

Expand

\frac{t ^{6} + 1}{t} : t^{5} + \frac{1}{t}

= \int_{1}^{e} t^{5} + \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^{5} d t + \int_{1}^{e} \frac{1}{t} d t

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

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

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

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What is the integral from 1 to e of ((t^6+1))/t ?
The integral from 1 to e of ((t^6+1))/t is (e^6-1)/6+1