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

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

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

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

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

Solution steps

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

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

Apply u-substitution

: \int \frac{( v - 1 ) ^{2}}{v} d v

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

Expand

\frac{( v - 1 ) ^{2}}{v} : v - 2 + \frac{1}{v}

= 3 \cdot \int v - 2 + \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

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

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 3 (\frac{v ^{2}}{2} - 2 v + ln ∣ v ∣)

Substitute back

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

Add a constant to the solution

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

\frac{d y}{d x} = x e^{2 y}

\frac{1}{( s - 1 ) \cdot ( s - 2 )}

\frac{d}{d x} (6 arcsin (x^{3}))

\frac{d}{d x} ((1 + cos (2 x))^{2})

\frac{d}{d x} (\frac{x ^{2} - 2 x}{x})

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