What is the integral from 0 to 2 of x^3sqrt(1+9x^4) ?

The integral from 0 to 2 of x^3sqrt(1+9x^4) is (145sqrt(145)-1)/(54)

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integral from 0 to 2 of x^3sqrt(1+9x^4)

\int_{0}^{2} x^{3} 1 + 9 x^{4} d x

\frac{145 145 - 1}{54}

Solution steps

\int_{0}^{2} x^{3} 1 + 9 x^{4} d x

Apply u-substitution

: \int_{1}^{145} \frac{u}{36} d u

= \int_{1}^{145} \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{1}^{145} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Compute the boundaries:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{290 145}{3} - \frac{2}{3})

Simplify

= \frac{145 145 - 1}{54}

\frac{d}{d x} (8 arctan (x))

\int_{1}^{e} \frac{1}{x} d x

y = ln (\frac{1}{x x + 1})

x y^{^{'}} - 2 y - 16 = 0

y^{^{'}} = e^{5 y} sin (6 x)

What is the integral from 0 to 2 of x^3sqrt(1+9x^4) ?
The integral from 0 to 2 of x^3sqrt(1+9x^4) is (145sqrt(145)-1)/(54)