Exponentials/Logarithms (1 Viewer)

bouncing · Mar 21, 2010

Hey guys can i get some help with these questions:

integrate: xe^3x^2 dx
(so xe to the power of 3x squared)

find:
integral of (4x^2/x^3-1)dx

differentiate:
a) y=3^2x+3
b) y=logbase5(3-x)

thongetsu · Mar 21, 2010

1. 6x^2.e^3x^2
2. 2log(x^3-1)

bouncing · Mar 21, 2010

how did you get that?
could you please show working out or something?

blackratpoo · Mar 21, 2010

Im not sure about all of these as they are a little different to the ones you'll get in your average 2U exam.

$Im not so sure about the 1st one but here you go$\\ \int xe^{3x^{2}}\qquad = \qquad \frac{1}{6x}xe^{3x^{2}}+c\\ \\ $Second one$\\ \int \frac{4x^{2}}{x^{3}-1} \qquad = \qquad \frac{4}{3}\int \frac{3x^{2}}{x^{3}-1} \qquad = \qquad \frac{4}{3}ln(x^{3}-1)+c\\ \\ a)y=3^{2x+3}\\ y'=2.3^{2x+3}\\ y'=6^{2x+3}\\ \\ b) y=log_{5}(3-x) $Not sure, but use the change of base rule and then differentiate i think.$

bouncing · Mar 21, 2010

thanks blackratpoo :L

hmm but these are from 2u fitz

EDIT: SORRY NO THEY ARE NOT :\ some random book i had

- Alex - · Mar 21, 2010

For the last question

Differentiate
$\log_{5}(3-x)$\\Change of base;$\\y=\frac{\log_{e}(3-x)}{\log_{e}5}$\\Take log5 out as it is a constant and differentiate using the function of a function rule $\\\therefore \frac{1}{\log_{e}5}\cdot \frac{-1}{(3-x)}$\\\therefore y'=\frac{-1}{\log_{e}5(3-x)}$

Not sure if i am correct though, if u have the answer let me know :/

Exponentials/Logarithms (1 Viewer)

bouncing

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thongetsu

Where aren't I?

bouncing

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blackratpoo

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bouncing

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- Alex -

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