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

The integral from 0 to 3 of sqrt(x^4+1)x^3 is (82sqrt(82)-1)/6

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

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

\frac{82 82 - 1}{6}

Decimal Notation

123.59026 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{82} \frac{u}{4} d u

= \int_{1}^{82} \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_{1}^{82} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{82}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{82}

Compute the boundaries:

\frac{164 82}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{164 82}{3} - \frac{2}{3})

Simplify

= \frac{82 82 - 1}{6}

\int \frac{x ^{2} + 8 x + 1}{x ^{3} + 12 x ^{2} + 3 x + 6} d x

\int \frac{- 4 y - 1}{y + 1} d y

f (x) = x (3 x - 1)

\frac{d}{d x} ((3 x + 4)^{6})

\frac{d}{d x} (- e^{7 x})

What is the integral from 0 to 3 of sqrt(x^4+1)x^3 ?
The integral from 0 to 3 of sqrt(x^4+1)x^3 is (82sqrt(82)-1)/6