anonymous
  • anonymous
Integral of 1/(x+1)^2. This feels like it should be simple, but I can't put my finger on the answer...
Calculus1
  • 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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
What is your best shot on the problem ?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{w^n}~dw=\int\limits_{~}^{~}w^{-n}dw=\frac{1}{-n+1}w^{-n+1}+C}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{w^2}~dw=\int\limits_{~}^{~}w^{-2}dw=\frac{1}{-2+1}w^{-2+1}+C=-w^{-1}+C}\)

Looking for something else?

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

More answers

anonymous
  • anonymous
I was thinking of product ruling it with (x+1)^-2 as u and dx as dv... but it doesn't seem like that'll go anywhere. The natural log antiderivat thing might work, but I haven't done it with a denominator to a power before...
SolomonZelman
  • SolomonZelman
No, see what I wrote above? Do you agree with what I wrote?
SolomonZelman
  • SolomonZelman
When (x+1) is your variable that makes no difference because dx would =du if you set u=x+1, and thus you can just treat x+1 as a single variable, or if you like you can perform that u sub anyway
anonymous
  • anonymous
pellet, you're right. Goddamn substitution. Thanks.
SolomonZelman
  • SolomonZelman
Anytime \(!\)

Looking for something else?

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