What is the integral of 1/(1+\sqrt[3]{2x+1)} ?

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

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integral of 1/(1+\sqrt[3]{2x+1)}

Solution steps

Apply u-substitution

Take the constant out:

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

Apply u-substitution

: \int \frac{3 v ^{2}}{1 + v} d v

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 2 d w = 2 w

\int \frac{1}{w} d w = ln ∣ w ∣

= \frac{1}{2} \cdot 3 (\frac{w ^{2}}{2} - 2 w + ln ∣ w ∣)

Substitute back

Simplify

Add a constant to the solution

\int (x - 5) (3 x + 4) d x

\int (5 x^{3} + 6 x^{3} - 5) d x

\int (5 x + 2)^{7} d x

\int (2 t^{2} - 3)^{2} d t

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

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