What is the integral of 1/(x^{2/3)(8+x^{1/3})} ?

The integral of 1/(x^{2/3)(8+x^{1/3})} is 3ln|8+x^{1/3}|+C

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integral of 1/(x^{2/3)(8+x^{1/3})}

\int \frac{1}{x ^{\frac{2}{3}} ( 8 + x ^{\frac{1}{3}} )} d x

3 ln 8 + x^{\frac{1}{3}} + C

Solution steps

\int \frac{1}{x ^{\frac{2}{3}} ( 8 + x ^{\frac{1}{3}} )} d x

Apply u-substitution

: \int \frac{3}{u} d u

= \int \frac{3}{u} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 ln ∣ u ∣

Substitute back

u = 8 + x^{\frac{1}{3}}

= 3 ln 8 + x^{\frac{1}{3}}

Add a constant to the solution

= 3 ln 8 + x^{\frac{1}{3}} + C

y (θ) = ln (θ + 1) - e^{θ}

\frac{\partial}{\partial x} (ln (x + y) - 5 y)

\frac{d}{d x} ((x + 3)^{x})

\frac{d}{d x} (x^{5 e^{x}})

ln (\frac{x}{2} - 3), [7, 9]

What is the integral of 1/(x^{2/3)(8+x^{1/3})} ?
The integral of 1/(x^{2/3)(8+x^{1/3})} is 3ln|8+x^{1/3}|+C