Recent content by jrmed13

  1. J

    Improper Integral Problem (I just want to check my answer)

    Homework Statement integral (from 0 to 1) of (lnx)dx/(x^0.5) Homework Equations I did u-substitution and got the antiderivative to be 4ln(sqrt(x)) - 4sqrt(x) The Attempt at a Solution The answer that I got was that the limit of the antiderivative (as t approaches 0 from the...
  2. J

    Integrating Natural Log Function using Integration by Parts Method

    Integrating Natural Log Function using "Integration by Parts" Method Homework Statement The problem says to integrate ln(2x+1)dx Homework Equations I used u=ln(2x+1); du = 2dx/(2x+1); dv=dx; v=x The Attempt at a Solution So, I integrated it using that (above) 'dictionary' and I...
  3. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    Oh waiiiiiit never mind I found it in my reference page (integral of du/(x^2 + a^2) = blahblah. Cristo, thank you so much - you are amaaaaaaaaaazing and although I did cry over this problem for about ninety minutes, I have learned a bunch and I am going to memorize some of these integrals...
  4. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    I mean that integral instead of derivative
  5. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    Wait, what? I don't follow - I have never seen that before in Calculus I (I'm in Calc. II right now) And what about the 9?? That derivative is for x^2 + 1, not x^2 + 9
  6. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    sorry for my constant refrain of complaining, but I already tried that and couldn't figure out how to integrate (u^2 + 9)^-1... I know it involves something with ln(u^2 + 9), but I couldn't get that to work either!
  7. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    How?? I tried that million times and I couldn't get the right answer! Here's what I tried u=e^7x du=(7e^7x)dx so then the integrand (that's what the thing being integrated is called, right?) becomes (u/(u^2 + 9))*du/(7e^7x) that doesn't work!
  8. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    I know that.. but how do I work from there? The problem with "e" is that its derivative is different than integrating something like x^2
  9. J

    How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

    Homework Statement So, I have to integrate the following expression: (e^7x)/(e^(14x) + 9)dx Homework Equations We are doing the section on 'integrating by u-substitution' right now, so that might help in finding a solution... The Attempt at a Solution So, I tried a bunch of...
  10. J

    Test on monday (banked road problem)

    hm... Okay, so maybe I should make my question more clear... here goes... The problem: The Daytona International Speedway in Daytona Beach, FL, is famous for its races(blahblahblah)... Both of its courses feature four-story 31.0degree banked curves, with maximum radius of 316m. If the car...
  11. J

    Test on monday (banked road problem)

    URGENT! test on monday! (banked road problem) Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal...
  12. J

    Max Velocity for Banked Road Problem

    Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal acceleration by saying that (mv^2)/r =...
  13. J

    Solving Banked Road Problem with Car Speeds of 48 km/h

    Help! Okay, so I am doing a similar problem involving a car driving on a banked, circular track (theta=31degrees). I know that to find the centripetal acceleration, I am supposed to say that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am...
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