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

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

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

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

ln y^{3} + y + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = y^{3} + y

= ln y^{3} + y

Add a constant to the solution

= ln y^{3} + y + C

\int sin^{3} (x) d x

x \to - 2 lim (x (x - 1))

\frac{2000 x}{4 x + 75}

\int y + 1 d y

9 cos (x), 7 sin (x), 0, π

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