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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 Notation

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

\int_{1}^{2} arctan (x) d x

\int_{- 0.5}^{0.5} 8 x^{2} - 2 d x

\int_{0}^{0.8} 48000 x - 21600 d x

\int_{0}^{π} 2 - 2 cos (x) d x

\int_{1}^{x} sin (t) d t

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