Question

evaluate the integral.
\int \frac{3y^2 \\, dy}{y^3 - 14}

\int \frac{3y^2 \\, dy}{y^3 - 14} = \square

Explanation:

Step1: Use substitution method

Let \( u = y^3 - 14 \), then find the derivative of \( u \) with respect to \( y \).
\( \frac{du}{dy} = 3y^2 \), so \( du = 3y^2 dy \).

Step2: Substitute into the integral

The integral \( \int \frac{3y^2 dy}{y^3 - 14} \) becomes \( \int \frac{du}{u} \) (since \( 3y^2 dy = du \) and \( u = y^3 - 14 \)).

Step3: Integrate \( \int \frac{du}{u} \)

The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln|u| + C \), where \( C \) is the constant of integration.

Step4: Substitute back \( u = y^3 - 14 \)

We get \( \ln|y^3 - 14| + C \).

Answer:

\( \ln|y^3 - 14| + C \)