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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{( e ^{x} )}{2} d x

x \to \frac{1}{2} lim (\frac{x ^{2}}{2})

3 x^{2} - x + 1

f (x) = x^{2} + 6 (- 5.31)

\int \frac{1}{1 x} d x

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