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

The integral from 1 to 2 of x^2sqrt(x^3+1) is (54-4sqrt(2))/9

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

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

\frac{54 - 4 2}{9}

Decimal

5.37146 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{9} \frac{u}{3} d u

= \int_{2}^{9} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{2}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

Compute the boundaries:

18 - \frac{4 2}{3}

= \frac{1}{3} (18 - \frac{4 2}{3})

Simplify

= \frac{54 - 4 2}{9}

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What is the integral from 1 to 2 of x^2sqrt(x^3+1) ?
The integral from 1 to 2 of x^2sqrt(x^3+1) is (54-4sqrt(2))/9