What is the integral of (y^2)/(1+y^3) ?

The integral of (y^2)/(1+y^3) is 1/3 ln|1+y^3|+C

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integral of (y^2)/(1+y^3)

\int \frac{y ^{2}}{1 + y ^{3}} d y

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

Solution steps

\int \frac{y ^{2}}{1 + y ^{3}} d y

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + y^{3}

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

Add a constant to the solution

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

y = 9 e^{x} cos (x), (0, 9)

\frac{d}{d x} ((x + sin (x))^{5})

3 tan (x), at x = \frac{π}{3}

\frac{d}{d θ} (2 sin (θ) + 2 cos (θ))

(9.5) (21.1)

What is the integral of (y^2)/(1+y^3) ?
The integral of (y^2)/(1+y^3) is 1/3 ln|1+y^3|+C