- anonymous

the indefinite integral i am given is (e^x - 6x^2)/6... i need to evaluate it. please help!

- schrodinger

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- nowhereman

Do you know the integral of e^x and the power rule?

- anonymous

i assume the integral is e^x

- anonymous

i need to use substitution, but i dont know what to substitute... i've tried it a lot already

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## More answers

- nowhereman

You don't need to use any elevated integration techniques. If you want to prove that your result is right, just differentiate it and if you get the original integrand, you're fine.

- anonymous

what about the 6 in the denominator? how do i do that?
is it just 6x?

- nowhereman

No, that is just a factor of 1/6 which you can pull-out.

- anonymous

wow.. i got it! thanks! i have another one tho if u are willing to help?

- nowhereman

Sure :-)

- anonymous

(5-2xe^x)/x

- nowhereman

Pull that division into the difference and you should have two known integrals.

- anonymous

like (5/x)-((2xe^x)/x)?

- nowhereman

Yes, and reduce that x in the second term.

- anonymous

ok. then i would need substitution right?

- nowhereman

For what? You should know the integral of 1/x and e^x and that's all there is.

- anonymous

i dont know the 1/x one. is it natural log?

- nowhereman

Exactly, sometimes it is defined this way \[\ln x := \int_1^x\frac 1 y \mathrm d y\] but if you have defined it as the inverse function of e^x you could use the inverse-differentiation rule.

- anonymous

i got that one! thanks a lot. this is my last one... (x-2)/(x^2-4x+4)^3..... that requires substitution correct?

- nowhereman

Yes, a little one :-) but you should first observe that \[x^2 - 4x + 4 = (x-2)^2\]

- anonymous

so i can cancel one of them right away?

- nowhereman

yes

- anonymous

would it turn into (x-2)^6?

- anonymous

in the denominator

- nowhereman

Sure, that is just one of the exponentiation rules.

- anonymous

then after the cancellation i am left with 1/(x-2)^5?

- nowhereman

Right.

- anonymous

now i substitute u=x-2

- nowhereman

Yes, but as du/dx = 1 if you are a little more experienced you could just do it in your head.

- anonymous

i think it would be 1/((1/6)*(x-2)^6)

- nowhereman

Not quite, maybe you should write it is \[(x-2)^{-5}\] instead to see it better.

- anonymous

(-1/4)*(x-2)^-4?

- nowhereman

Yeah, good.

- anonymous

is this website free to use?

- nowhereman

It costs both of us our time, but you won't have to pay for it ;-)

- anonymous

so y do ppl like u go around and answer questions?

- nowhereman

I love math and helping other people.

- anonymous

it seems strange that there is no charge for this service. or breech of security

- nowhereman

I don't know what you mean by breech of security; but the idea of being is just that everyone can ask questions and give answers here. So if I had a question regarding my study, I could ask them too. Only that it seems there are a lot of high-school students here lately :-D

- anonymous

hmm... thats cool. i guess i'm a skeptic. but with quick responses like the ones u gave me i am becoming a believer

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