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

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

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

\int \frac{1}{x - 2 x ^{\frac{2}{3}} - 3 x ^{\frac{1}{3}}} d x

3 (\frac{1}{4} ln x^{\frac{1}{3}} + 1 + \frac{3}{4} ln x^{\frac{1}{3}} - 3) + C

Solution steps

\int \frac{1}{x - 2 x ^{\frac{2}{3}} - 3 x ^{\frac{1}{3}}} d x

Apply u-substitution

= \int \frac{3 u}{u ^{2} - 2 u - 3} 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} - 2 u - 3} d u

Take the partial fraction of

\frac{u}{u ^{2} - 2 u - 3} : \frac{1}{4 ( u + 1 )} + \frac{3}{4 ( u - 3 )}

= 3 \cdot \int \frac{1}{4 ( u + 1 )} + \frac{3}{4 ( u - 3 )} d u

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 \frac{1}{4 ( u + 1 )} d u + \int \frac{3}{4 ( u - 3 )} d u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{3}{4 ( u - 3 )} d u = \frac{3}{4} ln ∣ u - 3 ∣

= 3 (\frac{1}{4} ln ∣ u + 1 ∣ + \frac{3}{4} ln ∣ u - 3 ∣)

Substitute back

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

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

Add a constant to the solution

= 3 (\frac{1}{4} ln x^{\frac{1}{3}} + 1 + \frac{3}{4} ln x^{\frac{1}{3}} - 3) + C

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