What is the integral from 1 to 2 of x^3sqrt(x^4+5) ?

The integral from 1 to 2 of x^3sqrt(x^4+5) is (7sqrt(21)-2sqrt(6))/2

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integral from 1 to 2 of x^3sqrt(x^4+5)

\int_{1}^{2} x^{3} x^{4} + 5 d x

\frac{7 21 - 2 6}{2}

Decimal

13.58952 \dots

Solution steps

\int_{1}^{2} x^{3} x^{4} + 5 d x

Apply u-substitution

= \int_{6}^{21} \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_{6}^{21} u d u

Apply the Power Rule

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{6}^{21}

Compute the boundaries:

1421 - 46

= \frac{1}{4} (1421 - 46)

Simplify

= \frac{7 21 - 2 6}{2}

What is the integral from 1 to 2 of x^3sqrt(x^4+5) ?
The integral from 1 to 2 of x^3sqrt(x^4+5) is (7sqrt(21)-2sqrt(6))/2