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Math Help - Substitution Rule Integral Questions

  1. #1
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    Substitution Rule Integral Questions

    I have a few questions I'm unsure about, the first one is a indefinite integral:

    1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

    I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx

    The next two are definite

    2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

    For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19. The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.

    3.) antiderivative x/(x^2+1)ln(x^2+1)

    For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

    But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.
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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have a few questions I'm unsure about, the first one is a indefinite integral:

    1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

    I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx
    correct. but write your answer in terms of x and don't forget the +C

    The next two are definite

    2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

    For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19. The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.
    sorry, don't see your problem here, i got the same answer both ways. if you are going to back-substitute the x expression, there is no point in changing the limit in the first place.

    3.) antiderivative x/(x^2+1)ln(x^2+1)

    For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

    But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.
    you lost me. where'd 0.69 and 1.61 come from. how did the limits become 1 and 2?

    also, if i read your problem correctly (you should use parentheses), i think you mean \int \frac x{(x^2 + 1)\ln (x^2 + 1)}~dx as opposed to \int \frac x{x^2 + 1} \cdot \ln (x^2 + 1)~dx

    then, the integral is: \frac 12 \ln \bigg| \ln |x^2 + 1| \bigg| + C
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have a few questions I'm unsure about, the first one is a indefinite integral:

    1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

    I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx

    Mr F says: You can check your answer by differentiating it (and don't forget to add the 'C' to your answer).

    The next two are definite

    2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

    For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19.

    Mr F says: Now calculate this definite integral.

    The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.

    Mr F says: Why would you substitute back?? This is totally unnecessary. It wastes time and increases the probabilty of making an error and getting the wrong answer.

    3.) antiderivative x/(x^2+1)ln(x^2+1)

    For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

    But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.
    For questions 2.) and 3.) it will be easier to help you if you show all your working.
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  4. #4
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    For the third one, I'm getting the same antiderivative as you. When I compute it between 1 and 2, I get an answer of 0.46.

    For the second one, I got an answer of -1.31.
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  5. #5
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    For the third one, I'm getting the same antiderivative as you. When I compute it between 1 and 2, I get an answer of 0.46.

    For the second one, I got an answer of -1.31.
    The answer for the second one is positive not negative. And it's possible to give an exact answer: \frac{\sqrt{19} - \sqrt{3}}{2}.

    The answer for the third one, correct to two decimal places, is 0.42. Again, it's possible to give an exact answer: \frac{1}{2} \, \ln \frac{\ln 5}{\ln 2}. You're probably making a data entry error when attempting to get an approximate answer.
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