What is the integral from 0 to 1 of (\sqrt[4]{x}+1)^2 ?

The integral from 0 to 1 of (\sqrt[4]{x}+1)^2 is 49/15

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integral from 0 to 1 of (\sqrt[4]{x}+1)^2

\frac{49}{15}

Decimal Notation

3.26666 \dots

Solution steps

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int_{0}^{1} x d x = \frac{2}{3}

\int_{0}^{1} 1 d x = 1

= \frac{2}{3} + \frac{8}{5} + 1

Simplify

\frac{2}{3} + \frac{8}{5} + 1 : \frac{49}{15}

= \frac{49}{15}

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What is the integral from 0 to 1 of (\sqrt[4]{x}+1)^2 ?
The integral from 0 to 1 of (\sqrt[4]{x}+1)^2 is 49/15