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

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

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

\frac{3}{2} ln 1 + y^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

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

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

Simplify

= \frac{3}{2} ln 1 + y^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{2} ln 1 + y^{\frac{2}{3}} + C

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