What is the integral of 1/(x^{1/3)(x^{2/3}+4)} ?

The integral of 1/(x^{1/3)(x^{2/3}+4)} is 3/2 ln|x^{2/3}+4|+C

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integral of 1/(x^{1/3)(x^{2/3}+4)}

\int \frac{1}{x ^{\frac{1}{3}} ( x ^{\frac{2}{3}} + 4 )} d x

\frac{3}{2} ln x^{\frac{2}{3}} + 4 + C

Solution steps

\int \frac{1}{x ^{\frac{1}{3}} ( x ^{\frac{2}{3}} + 4 )} d x

Apply u-substitution

: \int \frac{3 u}{u ^{2} + 4} d u

= \int \frac{3 u}{u ^{2} + 4} d u

Take the constant out:

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

= 3 \cdot \int \frac{u}{u ^{2} + 4} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= 3 \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 3 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= 3 \cdot \frac{1}{2} ln (x^{\frac{1}{3}})^{2} + 4

Simplify

3 \cdot \frac{1}{2} ln (x^{\frac{1}{3}})^{2} + 4 : \frac{3}{2} ln x^{\frac{2}{3}} + 4

= \frac{3}{2} ln x^{\frac{2}{3}} + 4

Add a constant to the solution

= \frac{3}{2} ln x^{\frac{2}{3}} + 4 + C

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What is the integral of 1/(x^{1/3)(x^{2/3}+4)} ?
The integral of 1/(x^{1/3)(x^{2/3}+4)} is 3/2 ln|x^{2/3}+4|+C