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How To Integrate These?

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data

    FIRST
    ##\int x^2 \ln x \ dx##

    SECOND
    ##\int \frac{(x+1)^2}{x}dx##

    THIRD
    ##\int \frac{x-2}{x+3}dx##

    2. Relevant equations

    N/A

    3. The attempt at a solution
    I tried using the integration by part method but it doesn't work.

    For the FIRST problem:
    Setting ##u=x^2## or ##u=\ln x## is not solving the problem.

    For the SECOND problem:
    Setting ##u=x+1## or ##u=x## is not solving the problem.

    For the THIRD problem:
    Setting ##u=x-2## or ##u=x+3## is not solving the problem.
     
    Last edited: Dec 1, 2014
  2. jcsd
  3. Dec 1, 2014 #2

    Orodruin

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    You need to provide more details than this on your attempted solution. Show us what you did and what you got.
     
  4. Dec 1, 2014 #3
    For the FIRST problem:
    Setting ##u=x^2## or ##u=\ln x## is not solving the problem.

    For the SECOND problem:
    Setting ##u=x+1## or ##u=x## is not solving the problem.

    For the THIRD problem:
    Setting ##u=x-2## or ##u=x+3## is not solving the problem.
     
  5. Dec 1, 2014 #4

    ehild

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    What is u? show how did you proceed in the first case.
    How do you do the integration by parts?
     
  6. Dec 1, 2014 #5
    Give me some hints.
     
  7. Dec 1, 2014 #6

    pasmith

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    Differentiating [itex]\ln x[/itex] and integrating [itex]x^2[/itex] is the correct method. What did you get when you tried it?

    The remaining two problems do not require integration by parts. They can be done by direct integration after some algebraic manipulation of the integrand.
     
  8. Dec 1, 2014 #7
    Thank you

    This is the FIRST problem's solution done by me after getting a hint from you:
    let ##u=\ln x## then ##\frac{du}{dx}=\frac{1}{x}## or ##du=\frac{1}{x}dx##
    let ##dv=x^2## then ##v=\frac{1}{3}x^3##

    So ##\int x^2 \ln x \ dx = u.v-\int v.du##
    ##=\ln x . \frac{1}{3}x^3-\int\frac{1}{3}x^3(\frac{1}{x}dx)##
    ##=\ln x . \frac{1}{3}x^3-\frac{1}{3}\int\frac{x^3}{x}dx##
    ##=\ln x . \frac{1}{3}x^3-\frac{1}{3}\int x^2dx##
    ##=\ln x . \frac{1}{3}x^3-\frac{1}{3}[\frac{1}{3}x^3+c]##
    ##=\ln x . \frac{1}{3}x^3-\frac{1}{9}x^3+c##
    CMIIW.

    How to do the algebraic manipulation? I don't understand. Give some hints again for the SECOND and THIRD problem.
     
  9. Dec 1, 2014 #8

    ehild

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    expand the square in the second problem.
     
  10. Dec 1, 2014 #9
    You're right!

    ##\int\frac{(x+1)^2}{x}dx##
    ##=\int\frac{x^2+2x+1}{x}dx##
    ##=\int\frac{x^2}{x}dx+\int\frac{2x}{x}dx+\int\frac{1}{x}dx##
    ##=\int x \ dx+2\int dx+\int\frac{1}{x}dx##
    ##=\frac{1}{2}x^2+2x+\ln x+c##

    Is it correct?

    One more problem to solve.

    Give me a hint.
     
  11. Dec 1, 2014 #10

    ehild

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    Write the numerator as the sum (x+3) -5.

    You have the integral
    ##
    \int \frac{(x+3)-5}{x+3}dx=\int 1-\frac{5}{x+3}dx
    ##
     
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