Evaluate the indefinite integral. ∫12x2+24xx3+3x2+2

Step 1
Consider the provided indefinite integral,
Simplify the given indefinite integral as follows,
${\displaystyle \int \frac{12x^2+24x}{x^3+3x^2+2}}$
Apply u-substotution,
let ${\displaystyle u=x^3+3x^2+2}$
${\displaystyle \frac{du}{dx}=3x^2+6x}$
${\displaystyle du=(3x^2+6x)dx}$
multiply 4 in both the sides,
${\displaystyle 4du=(12x^2+24x)dx}$
Step 2
Now, the given indefinite integral is written as,
${\displaystyle \int \frac{12x^2+24x}{x^3+3x^2+2}dx=\int \frac{4du}{u}}$
${\displaystyle =4\int \frac{du}{u}}$
${\displaystyle =4\ln \left|u\right|+C}$
Substitute back, ${\displaystyle u=x^3+3x^2+2}$
${\displaystyle =4\ln |x^3+3x^2+2|+C}$
Thus, ${\displaystyle \int \frac{12x^2+24x}{x^3+3x^2+2}dx=4\ln |x^3+3x^2+2|+C}$

${\displaystyle \int \frac{12\cdot x^2+24x}{x^3+3\cdot x^2+2}dx}$
Lets

$\begin{array}{}\int \frac{12x^2+24x}{x^3+3x^2+2}\\ =4\int \frac{1}{u}du\\ \int \frac{1}{u}du\\ =\ln (u)\\ 4\int \frac{1}{u}du\\ =\ln (u)\\ 4\int \frac{1}{u}du\\ =4\ln (u)\\ =4\ln (x^3+3x^2+2)\\ \int \frac{12x^2+24x}{x^3+3x^2+2}dx\\ =4\ln (|x^3+3x^2+2|)+C\end{array}$