anonymous
  • anonymous
finding the anti derivatives by substitution. 2/(1-x)^2dx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1378261160747:dw|
anonymous
  • anonymous
oh parenthesis around the (1-x)^2
anonymous
  • anonymous
I have u=1-x du=d(1-x) du=-1 dx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
good, from there, multiply both sides by -1 in your last step
zepdrix
  • zepdrix
\[\Large dx=-du \qquad\text{right?}\]
anonymous
  • anonymous
ohh. yes.
anonymous
  • anonymous
i dont get what to do after you sub in the u. you end up with 2/u^2
zepdrix
  • zepdrix
\[\Large \int\limits \frac{2}{u^2}\left(-du\right)\]
zepdrix
  • zepdrix
Recall rules of exponents:\[\Large \frac{1}{x^2}\quad=\quad x^{-2}\]
zepdrix
  • zepdrix
Understand how to proceed from there? :)
anonymous
  • anonymous
uhh do you end up with 2u^-2?
zepdrix
  • zepdrix
Before integrating? yes that looks good. \[\Large -\int\limits 2u^{-2}\;du\] Don't forget about the negative from before! :)
zepdrix
  • zepdrix
You could even write it like this if you want,\[\Large -2\int\limits\limits u^{-2}\;du\]And simply apply the Power Rule for Integration from here :O yes?
anonymous
  • anonymous
4/(1-x)^3?
zepdrix
  • zepdrix
\[\Large \int\limits u^{-2}\;du \quad=\quad \frac{1}{-2+1}u^{-2+1}\]Woops! It looks like you subtracted one from the exponent :O like with derivatives.
anonymous
  • anonymous
what did you do with the -2?
anonymous
  • anonymous
i have -2 integral -2u^-1
zepdrix
  • zepdrix
I wasn't showing it with the -2. That was just an example.
anonymous
  • anonymous
i keep getting 4/(x-1)
zepdrix
  • zepdrix
\[\Large -2\int\limits u^{-2}\;du \quad=\quad -2(-u^{-1})\]
zepdrix
  • zepdrix
Incease power by one ( `goes from -2 to -1` ) then divide by the new exponent ( `-1` )
anonymous
  • anonymous
okay! thanks. its -2/(x-1)
zepdrix
  • zepdrix
ya sounds right! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.