Algotssleeddynf

2021-12-29

Evaluate the indefinite integral.
$\int \frac{12{x}^{2}+24x}{{x}^{3}+3{x}^{2}+2}$

Stella Calderon

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

Ella Williams

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

karton

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

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