Results 1 to 6 of 6

Math Help - Integration by Parts

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    Integration by Parts

    Hello guys, I've got a math quiz tomorrow. My professor only uses previous homework problems that he assigns so I'm just going over all of the problems in the book that he assigned, lol. I'm having trouble with this question:

    For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.




    <br />
{\int_{}\frac{lnx}{x^2}} dx<br />



    So... find the antiderivitive of: (lnx/x^2) dx ... u = lnx dv = (x^-2)dx

    I get this far:
    u = lnx
    du/dx = 1/x
    du = 1/x dx

    dv = x^-2dx
    dv/dx = x^-2
    v = x^-1/1

    So... u * v - integral (v * du)

    lnx * x^-1/1 - integral (x^-1/1 * 1/x) dx
    Last edited by larson; February 24th 2008 at 11:12 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by larson View Post
    For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.

    <br />
{\int_{}\frac{lnx}{x^2}} dx<br />
    Let f(x)=\frac{\ln x}x\implies f'(x)=\frac{1-\ln x}{x^2}.

    After integration we have \frac{{\ln x}}<br />
{x} + k_1  =  - \frac{1}<br />
{x} - \int {\frac{{\ln x}}<br />
{{x^2 }}\,dx} \,\therefore \,\int {\frac{{\ln x}}<br />
{{x^2 }}\,dx}  =  - \frac{1}<br />
{x} - \frac{{\ln x}}<br />
{x} + k.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Krizalid View Post
    Let f(x)=\frac{\ln x}x\implies f'(x)=\frac{1-\ln x}{x^2}.

    After integration we have \frac{{\ln x}}<br />
{x} + k_1  =  - \frac{1}<br />
{x} - \int {\frac{{\ln x}}<br />
{{x^2 }}\,dx} \,\therefore \,\int {\frac{{\ln x}}<br />
{{x^2 }}\,dx}  =  - \frac{1}<br />
{x} - \frac{{\ln x}}<br />
{x} + k.
    hey, I'm still a little confused... you didn't use integration by parts did you?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by larson View Post
    Hello guys, I've got a math quiz tomorrow. My professor only uses previous homework problems that he assigns so I'm just going over all of the problems in the book that he assigned, lol. I'm having trouble with this question:

    For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.




    <br />
{\int_{}\frac{lnx}{x^2}} dx<br />



    So... find the antiderivitive of: (lnx/x^2) dx ... u = lnx dv = (x^-2)dx

    I get this far:
    u = lnx
    du/dx = 1/x
    du = 1/x dx

    dv = x^-2dx
    dv/dx = x^-2
    v = x^-1/1

    So... u * v - integral (v * du)

    lnx * x^-1/1 - integral (x^-1/1 * 1/x) dx
    Krizalid's way is nice and easy, as usual (oy!) -- no, Krizalid did not use integration by parts. he literally reversed the derivative by integrating both sides of the derivative equation.

    but if you insist on doing integration by parts, i'd do a substitution first

    Let t = \ln x \implies e^t = x

    \Rightarrow dt = \frac 1x~dx

    So our integral becomes:

    \int te^{-t}~dt

    which is a lot less intimidating to do integration by parts on


    of course, your original approach may work, i didn't check it. this is just the way i saw
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Or integrate by parts as it's written:

    By setting u=\ln x\implies du=\frac1x\,dx & dv=x^{-2}\,dx\implies v=-\frac1x, your integral becomes

    \int {\frac{{\ln x}}<br />
{{x^2 }}\,dx}  =  - \frac{{\ln x}}<br />
{x} + \int {\frac{1}<br />
{{x^2 }}\,dx}, the rest follows.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Krizalid View Post
    Or integrate by parts as it's written:

    By setting u=\ln x\implies du=\frac1x\,dx & dv=x^{-2}\,dx\implies v=-\frac1x, your integral becomes

    \int {\frac{{\ln x}}<br />
{{x^2 }}\,dx}  =  - \frac{{\ln x}}<br />
{x} + \int {\frac{1}<br />
{{x^2 }}\,dx}, the rest follows.
    Now I understand, thank you both Krizalid and Jhevon!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum