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

The integral from-1 to 0 of 1/((x+1)^{1/3)} is 3/2

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integral from-1 to 0 of 1/((x+1)^{1/3)}

\int_{- 1}^{0} \frac{1}{( x + 1 ) ^{\frac{1}{3}}} d x

\frac{3}{2}

Decimal

1.5

Solution steps

\int_{- 1}^{0} \frac{1}{( x + 1 ) ^{\frac{1}{3}}} d x

Apply u-substitution

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

Apply the Power Rule

= [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{1}

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

What is the integral from-1 to 0 of 1/((x+1)^{1/3)} ?
The integral from-1 to 0 of 1/((x+1)^{1/3)} is 3/2