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

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

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

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

3 (ln (2) - \frac{1}{2})

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

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}^{2} v d v - \int_{1}^{2} 2 d v + \int_{1}^{2} \frac{1}{v} d v)

\int_{1}^{2} v d v = \frac{3}{2}

\int_{1}^{2} 2 d v = 2

\int_{1}^{2} \frac{1}{v} d v = ln (2)

= 3 (\frac{3}{2} - 2 + ln (2))

Combine the fractions

- 2 + \frac{3}{2} : - \frac{1}{2}

= 3 (ln (2) - \frac{1}{2})

\frac{d ^{2}}{d x ^{2}} (sin (x) + cos (x))

7 cos (π x), 8 x^{2} - 2, - 0.5, 0.5

\int e^{\frac{x}{6}} d x

\frac{2 t}{2 + t}

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