What is the integral from 0 to 6 of x^3sqrt(x^2+36) ?

The integral from 0 to 6 of x^3sqrt(x^2+36) is (31104sqrt(2)-7776)/5-36(144sqrt(2)-72)

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integral from 0 to 6 of x^3sqrt(x^2+36)

\int_{0}^{6} x^{3} x^{2} + 36 d x

\frac{31104 2 - 7776}{5} - 36 (1442 - 72)

Decimal Notation

2503.05662 \dots

Solution steps

\int_{0}^{6} x^{3} x^{2} + 36 d x

Apply u-substitution

: \int_{6}^{62} (u^{2} - 36) u^{2} d u

= \int_{6}^{62} (u^{2} - 36) u^{2} d u

Expand

(u^{2} - 36) u^{2} : u^{4} - 36 u^{2}

= \int_{6}^{62} u^{4} - 36 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_{6}^{62} u^{4} d u - \int_{6}^{62} 36 u^{2} d u

\int_{6}^{62} u^{4} d u = \frac{31104 2 - 7776}{5}

\int_{6}^{62} 36 u^{2} d u = 36 (1442 - 72)

= \frac{31104 2 - 7776}{5} - 36 (1442 - 72)

x \to - 3 - lim (\frac{( x + 2 )}{x + 3})

x (e^{\frac{y}{x}} - 1) y^{^{'}} = e^{\frac{y}{x}} (y - x), y (1) = 0

\frac{d}{d x} ((x^{2} + y^{2}) ln (x + y))

\frac{d}{d x} (\frac{x + 3}{x - 5})

\frac{\partial}{\partial y} (\frac{1}{2} ln (x^{2} + y^{2}))

What is the integral from 0 to 6 of x^3sqrt(x^2+36) ?
The integral from 0 to 6 of x^3sqrt(x^2+36) is (31104sqrt(2)-7776)/5-36(144sqrt(2)-72)