What is the integral from 1/4 to 7 of 9/(\sqrt[3]{4x-1)} ?

The integral from 1/4 to 7 of 9/(\sqrt[3]{4x-1)} is 243/8

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integral from 1/4 to 7 of 9/(\sqrt[3]{4x-1)}

\frac{243}{8}

Decimal Notation

30.375

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{27}

= 9 \cdot \frac{1}{4} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{27}

Simplify

9 \cdot \frac{1}{4} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{27} : \frac{9}{4} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{27}

= \frac{9}{4} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{27}

Compute the boundaries:

\frac{27}{2}

= \frac{9}{4} \cdot \frac{27}{2}

Simplify

= \frac{243}{8}

f (x) = - \frac{6}{( 1 + 3 x ) ^{2}}

\frac{d}{d x} (ln (\frac{( 2 )}{3 - x ^{2}}))

\frac{3}{t ^{3}}

n = 0 \sum \infty \frac{x ^{2 n}}{n !}

\frac{d}{d x} (\frac{e ^{x}}{x cos ( x )})

What is the integral from 1/4 to 7 of 9/(\sqrt[3]{4x-1)} ?
The integral from 1/4 to 7 of 9/(\sqrt[3]{4x-1)} is 243/8