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

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

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

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

\frac{17 17 - 1}{3}

Decimal Notation

23.03093 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{17} u^{2} d u

= \int_{1}^{17} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{17 17}{3} - \frac{1}{3}

= \frac{17 17}{3} - \frac{1}{3}

Simplify

= \frac{17 17 - 1}{3}

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