What is the integral from 0 to 3 of x^3sqrt(x^2+9) ?

The integral from 0 to 3 of x^3sqrt(x^2+9) is (972sqrt(2)-243)/5-9(18sqrt(2)-9)

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integral from 0 to 3 of x^3sqrt(x^2+9)

\int_{0}^{3} x^{3} x^{2} + 9 d x

\frac{972 2 - 243}{5} - 9 (182 - 9)

Decimal Notation

78.22051 \dots

Solution steps

\int_{0}^{3} x^{3} x^{2} + 9 d x

Apply u-substitution

: \int_{3}^{32} (u^{2} - 9) u^{2} d u

= \int_{3}^{32} (u^{2} - 9) u^{2} d u

Expand

(u^{2} - 9) u^{2} : u^{4} - 9 u^{2}

= \int_{3}^{32} u^{4} - 9 u^{2} d u

Apply the Sum Rule:

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

= \int_{3}^{32} u^{4} d u - \int_{3}^{32} 9 u^{2} d u

\int_{3}^{32} u^{4} d u = \frac{972 2 - 243}{5}

\int_{3}^{32} 9 u^{2} d u = 9 (182 - 9)

= \frac{972 2 - 243}{5} - 9 (182 - 9)

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What is the integral from 0 to 3 of x^3sqrt(x^2+9) ?
The integral from 0 to 3 of x^3sqrt(x^2+9) is (972sqrt(2)-243)/5-9(18sqrt(2)-9)