What is the integral from 0 to 2 of x^3sqrt(x^4+1) ?

The integral from 0 to 2 of x^3sqrt(x^4+1) is (17sqrt(17)-1)/6

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integral from 0 to 2 of x^3sqrt(x^4+1)

\int_{0}^{2} x^{3} x^{4} + 1 d x

\frac{17 17 - 1}{6}

Decimal

11.51546 \dots

Solution steps

\int_{0}^{2} x^{3} x^{4} + 1 d x

Apply u-substitution

: \int_{1}^{17} \frac{u}{4} d u

= \int_{1}^{17} \frac{u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{1}^{17} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{17 17 - 1}{6}

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What is the integral from 0 to 2 of x^3sqrt(x^4+1) ?
The integral from 0 to 2 of x^3sqrt(x^4+1) is (17sqrt(17)-1)/6