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Homework Help: Need help checking integration answers

  1. Mar 20, 2005 #1
    I just want someone to check since I only learnt this 'Integration using partial fractions' method today and I'm not sure I got it right...

    First question:
    [tex]\int \frac {dx}{x^2-a^2}[/tex] = [tex]\int \frac {dx}{(x+a)(x-a)}[/tex]

    let [tex] \frac {1}{(x+a)(x-a)} \equiv \frac {A}{x+a} + \frac {B}{x-a}[/tex]

    [tex]\equiv \frac {A(x-a)+B(x+a)}{(x+a)(x-a)}[/tex]

    so [tex] 1 \equiv (A+B)x+a(-A+B)[/tex]

    equate coefficients to find:
    [tex]B=\frac{1}{2a}[/tex] and [tex]A=\frac{-1}{2a}[/tex]

    substituting A and B back:
    [tex]\int \frac {dx}{(x+a)(x-a)}[/tex] = -[tex]\int \frac {x+a}{2a} dx + \int \frac {x-a}{2a} dx[/tex]

    I took the 1/2a out and integrated and got:
    [tex]\frac {-1}{2a} \frac {x^2}{2} + ax + \frac {1}{2a} \frac {x^2}{2} - ax + c[/tex]

    which simplifies to c :bugeye: which looks extremely strange to me...


    Second question:
    [tex]\int \frac {dx}{16x^4-1} = \int \frac {dx}{(4x^2+1)(2x+1)(2x-1)}[/tex]

    let [tex]\frac {1}{16x^4-1} \equiv \frac {Ax+B}{4x^2+1} + \frac {C}{2x+1} + \frac {D}{2x-1}[/tex]

    [tex]\equiv \frac {(Ax+B)(4x^2-1)+C(2x-1)(4x^2+1)+D(2x+1)(4x^2+1)}{16x^4-1}[/tex]

    [tex]1 \equiv x^3(4A+8C+8D)+x^2(4B-4C+4D)+x(-A+2C+2D)+D-C-B[/tex]

    equating coffecients to get:
    4A+8C+8D=0
    4B-4C+4D=0
    -A+2C+2D=0
    D-C-B=1

    I then found:
    A=0, B=-1/2, C=-1/4 and D=1/4

    substituting back into the integral:
    [tex]\int \frac {dx}{16x^4-1} = \frac {-1}{2} \int 4x^2+1 dx - \frac {1}{4} \int 2x+1 dx + \frac {1}{4} \int 2x+1[/tex]
    = [tex]\frac{-4x^3}{6}+x-\frac{x^2}{4}+x+\frac{x^2}{4}-x+c[/tex]
    = [tex]\frac {-2x^3}{3}+x+c[/tex]


    Last question:
    [tex]\int \frac {2}{(x^2+x+2)(x+1)}dx[/tex]

    let [tex]\frac {2}{(x^2+x+2)(x+1)} \equiv \frac {Ax+B}{(x+\frac {1}{2})^2 + \frac {7}{4}} + \frac {C}{x+1}[/tex]

    [tex]\equiv \frac {(Ax+B)(x+1)+C((x+\frac {1}{2})^2+\frac {7}{4})}{(x^2+x+2)(x+1)}[/tex]

    [tex]2\equiv x^2(A+C)+x(A+B+C)+B+2C[/tex]

    equating coefficients:
    A+C=0
    A+B+C=0
    B+2C=2

    so: A=-1, B=0, C=1

    putting it back into the integral:
    [tex]\int \frac {2}{(x^2+x+2)(x+1)}dx = -\int \frac {x}{(x+\frac {1}{2})^2+\frac {7}{4}} dx + \int \frac {1}{x+1} dx[/tex]

    I integrated the first part using the 'Integration by parts' method and then simplified to get:

    [tex]\frac {-x}{\sqrt {\frac{7}{4}}} \arctan (\frac {x+\frac{1}{2}}{\sqrt {7}}) - \tan (\frac {2x+1}{\sqrt{7}}) + ln(x+1) + c[/tex]

    Thanks in advance :)
     
  2. jcsd
  3. Mar 21, 2005 #2
    For #1 and #2, I want you to look very hard at the part where you subbed A and B back in and see if you see anything wrong (are you sure all those functions of x should be in the numerator?).

    I'll look at 3 in a minute~
     
  4. Mar 21, 2005 #3
    3 looks fine from the work you have down, but your answer is wrong. Check over your integration by parts~
     
  5. Mar 21, 2005 #4
    Oops....

    is #1
    [tex]\frac {-1}{2a} ln(2ax+2a^2) + \frac {1}{2a} ln(2ax-2a^2) + c
    ?

    for #2
    [tex]\frac {-1}{2} \int \frac {1}{4x^2+1} dx - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1)[/tex]
    I don't know how to integrate [tex]\int \frac {1}{4x^2+1} dx[/tex] ?

    #3...I got to:
    [tex]\frac {-x}{2(x^2+x+2)} - \int \frac {1}{2x^4+8x^2+8x+4}[/tex]
    but I don't know what to do after that...
     
  6. Mar 21, 2005 #5
    sorry, sorry, I mucked up the reply....

    it's supposed to be...



    Oops....

    is #1
    [tex]\frac {-1}{2a} ln(2ax+2a^2) + \frac {1}{2a} ln(2ax-2a^2) + c[/tex]
    ?

    for #2
    [tex]\frac {-1}{2} \int \frac {1}{4x^2+1} dx - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1)[/tex]
    I don't know how to integrate [tex]\int \frac {1}{4x^2+1} dx[/tex] though?

    #3...I got to:
    [tex]\frac {-x}{2(x^2+x+2)} - \int \frac {1}{2x^4+8x^2+8x+4}[/tex]
    but I don't know what to do after that...
     
  7. Mar 21, 2005 #6

    dextercioby

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    Homework Helper

    [tex] \int \frac{dx}{4x^{2}+1}=\frac{1}{2}\int \frac{d(2x)}{(2x)^{2}+1}=\frac{1}{2}\arctan 2x +C [/tex]

    Okay...?
     
  8. Mar 21, 2005 #7
    so #2 becomes
    [tex]\frac {-1}{4} \arctan(2x) - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1) + c[/tex] ?
     
  9. Mar 21, 2005 #8
    #1 looks right. You can simplify it to

    [tex]\frac{1}{2a} \ln \left(\frac{x-a}{x+a} \right) + C[/tex]

    too. Do you see how?

    #2 is perfect.

    For 3, note that

    [tex]\int \frac{2}{(x^2+x+2)(x+1)} \ dx = -\int \frac{x}{\left(x+\frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \int \frac{dx}{x+1}[/tex]

    [tex] = -\int \frac{x + \frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \int \frac{\frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \ln (x+1)[/tex]

    [tex] = -\int \frac{x + \frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \frac{1}{2\sqrt{7}}\arctan \left( \frac{2x+1}{\sqrt{7}} \right) + \ln (x+1)[/tex]

    and see if the substitution [itex] u = (x+ 1/2)^2[/itex] will help to evaluate the first integral.
     
    Last edited: Mar 21, 2005
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