What is the integral of (x^{1/3})/(x^{4/3)+4} ?

The integral of (x^{1/3})/(x^{4/3)+4} is 3/4 ln|x^{4/3}+4|+C

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integral of (x^{1/3})/(x^{4/3)+4}

\int \frac{x ^{\frac{1}{3}}}{x ^{\frac{4}{3}} + 4} d x

\frac{3}{4} ln x^{\frac{4}{3}} + 4 + C

Solution steps

\int \frac{x ^{\frac{1}{3}}}{x ^{\frac{4}{3}} + 4} d x

Apply u-substitution

: \int \frac{3}{4 u} d u

= \int \frac{3}{4 u} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{3}{4} ln ∣ u ∣

Substitute back

u = x^{\frac{4}{3}} + 4

= \frac{3}{4} ln x^{\frac{4}{3}} + 4

Add a constant to the solution

= \frac{3}{4} ln x^{\frac{4}{3}} + 4 + C

\frac{d y}{d x} + 4 y = \frac{4}{3}

\frac{\partial}{\partial x} (2 x e^{y} - x cos (x y) + 2 y)

y^{^{''}} + 45 y^{^{'}} + 20 y = 0

\int x x^{4} - 1 d x

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

What is the integral of (x^{1/3})/(x^{4/3)+4} ?
The integral of (x^{1/3})/(x^{4/3)+4} is 3/4 ln|x^{4/3}+4|+C