What is the integral of x/(x^{2/3)+x^{8/6}} ?

The integral of x/(x^{2/3)+x^{8/6}} is 3/2 (x^{2/3}+1-ln|x^{2/3}+1|)+C

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integral of x/(x^{2/3)+x^{8/6}}

\int \frac{x}{x ^{\frac{2}{3}} + x ^{\frac{8}{6}}} d x

\frac{3}{2} (x^{\frac{2}{3}} + 1 - ln x^{\frac{2}{3}} + 1) + C

Solution steps

\int \frac{x}{x ^{\frac{2}{3}} + x ^{\frac{8}{6}}} d x

Apply u-substitution

: \int \frac{3 u}{2 ( u + 1 )} d u

= \int \frac{3 u}{2 ( u + 1 )} d u

Take the constant out:

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

= \frac{3}{2} \cdot \int \frac{u}{u + 1} d u

Apply u-substitution

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

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

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{3}{2} \cdot \int 1 - \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

= \frac{3}{2} (\int 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{3}{2} (v - ln ∣ v ∣)

Substitute back

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

Add a constant to the solution

= \frac{3}{2} (x^{\frac{2}{3}} + 1 - ln x^{\frac{2}{3}} + 1) + C

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What is the integral of x/(x^{2/3)+x^{8/6}} ?
The integral of x/(x^{2/3)+x^{8/6}} is 3/2 (x^{2/3}+1-ln|x^{2/3}+1|)+C