Integral $\int_{}^{} (9x^2+6)/(3x^3+6x) dx$
I understand to use the chain rule but,
i get ln(3x^3+6x) but the 9x^2+6 has me bit confused....hmm
Hi Bee-Bee, learn to use this site.
You inadvertently entered "(9x^2+6)/(3x^3+6x^2)" into WolframAlpha, and that was what was last sitting there when I visited it.
It's really "(9x^2+6)/(3x^3+6x)."
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One of the forms of the answer is $\displaystyle ln|3x^3 + 6x| + C $, or you may see it as $\displaystyle log|3x^3 + 6x| + C $.
And, there are still other ways of writing the answer.
Regardless, it's the logarithm with the base e.
The graph of the function has a vertical asymptote of x = 0.