What is the integral of (12y^2)/(4y^3+1) ?

The integral of (12y^2)/(4y^3+1) is ln|4y^3+1|+C

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integral of (12y^2)/(4y^3+1)

\int \frac{12 y ^{2}}{4 y ^{3} + 1} d y

ln 4 y^{3} + 1 + C

Solution steps

\int \frac{12 y ^{2}}{4 y ^{3} + 1} d y

Take the constant out:

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

= 12 \cdot \int \frac{y ^{2}}{4 y ^{3} + 1} d y

Apply u-substitution

: \int \frac{1}{12 u} d u

= 12 \cdot \int \frac{1}{12 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 12 \cdot \frac{1}{12} ln ∣ u ∣

Substitute back

u = 4 y^{3} + 1

= 12 \cdot \frac{1}{12} ln 4 y^{3} + 1

Simplify

12 \cdot \frac{1}{12} ln 4 y^{3} + 1 : ln 4 y^{3} + 1

= ln 4 y^{3} + 1

Add a constant to the solution

= ln 4 y^{3} + 1 + C

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What is the integral of (12y^2)/(4y^3+1) ?
The integral of (12y^2)/(4y^3+1) is ln|4y^3+1|+C