What is the integral from 0 to 8 of 1/(v^{1/3)+1} ?

The integral from 0 to 8 of 1/(v^{1/3)+1} is 3ln(3)

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integral from 0 to 8 of 1/(v^{1/3)+1}

\int_{0}^{8} \frac{1}{v ^{\frac{1}{3}} + 1} d v

3 ln (3)

Decimal Notation

3.29583 \dots

Solution steps

\int_{0}^{8} \frac{1}{v ^{\frac{1}{3}} + 1} d v

Apply u-substitution

: \int_{0}^{2} \frac{3 u ^{2}}{u + 1} d u

= \int_{0}^{2} \frac{3 u ^{2}}{u + 1} d u

Take the constant out:

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

= 3 \cdot \int_{0}^{2} \frac{u ^{2}}{u + 1} d u

Apply u-substitution

: \int_{1}^{3} \frac{( w - 1 ) ^{2}}{w} d w

= 3 \cdot \int_{1}^{3} \frac{( w - 1 ) ^{2}}{w} d w

Expand

\frac{( w - 1 ) ^{2}}{w} : w - 2 + \frac{1}{w}

= 3 \cdot \int_{1}^{3} w - 2 + \frac{1}{w} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int_{1}^{3} w d w - \int_{1}^{3} 2 d w + \int_{1}^{3} \frac{1}{w} d w)

\int_{1}^{3} w d w = 4

\int_{1}^{3} 2 d w = 4

\int_{1}^{3} \frac{1}{w} d w = ln (3)

= 3 (4 - 4 + ln (3))

Simplify

= 3 ln (3)

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What is the integral from 0 to 8 of 1/(v^{1/3)+1} ?
The integral from 0 to 8 of 1/(v^{1/3)+1} is 3ln(3)