What is the integral from 0 to 5 of x^3sqrt(1+9x^4) ?

The integral from 0 to 5 of x^3sqrt(1+9x^4) is (5626sqrt(5626)-1)/(54)

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integral from 0 to 5 of x^3sqrt(1+9x^4)

\int_{0}^{5} x^{3} 1 + 9 x^{4} d x

\frac{5626 5626 - 1}{54}

Decimal Notation

7814.56490 \dots

Solution steps

\int_{0}^{5} x^{3} 1 + 9 x^{4} d x

Apply u-substitution

: \int_{1}^{5626} \frac{u}{36} d u

= \int_{1}^{5626} \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}^{5626} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5626}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5626}

Compute the boundaries:

\frac{11252 5626}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{11252 5626}{3} - \frac{2}{3})

Simplify

= \frac{5626 5626 - 1}{54}

\int \frac{4 x + 2}{e ^{x}} d x

\int \frac{1}{cos ^{2} ( t )} d t

f (x) = 2 x^{3} + x

\frac{d y}{d x} + y^{7} x + 3 y = 0

y^{^{'}} = \frac{x e ^{x}}{2 y}

What is the integral from 0 to 5 of x^3sqrt(1+9x^4) ?
The integral from 0 to 5 of x^3sqrt(1+9x^4) is (5626sqrt(5626)-1)/(54)