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Thread: Integration Question

  1. #1
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    Integration Question

    Hi
    I need help on the following questions:

    1)$\displaystyle \int \frac{x^2}{6x^2-7x+2}$
    This what i have done:

    do long division i get: $\displaystyle 1 - \frac{5x^2-7x+2}{6x^2-7x+2}$

    $\displaystyle \int 1dx \int \frac{5x^2-7x+2}{6x^2-7x+2}$

    do partial fractions we get:

    $\displaystyle \frac{-3}{2x-1} - \frac{1}{(3x-2)}$

    final solution i got is:
    $\displaystyle x - 3ln|2x-1| - ln|3x-2|$

    2)$\displaystyle \int \frac{x-1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

    $\displaystyle u=3x^2+4x+2 du=6x+4$

    $\displaystyle \frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

    final solution i got is:
    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$

    3)$\displaystyle \int \frac{3}{(x-+1)^2)(2x-3)}$
    This what i have done:

    split into partial fractions, after i get:

    $\displaystyle A=\frac{3}{5}$ $\displaystyle B=\frac{6}{25}$ $\displaystyle C=\frac{-12}{25}
    $
    $\displaystyle \int \frac{A}{(x-1)^2} + \frac{B}{(x+1)} + \frac{C}{(2x-3)} =
    \frac{3}{5(x-1)^2} + \frac{6}{25(x+1)} + \frac{-12}{25(2x-3)}$

    final solution i got is:

    $\displaystyle \frac{3}{5(x-1)^2} + \frac{6}{25}ln|x+1| - \frac{24}{25}ln|2x-3|$

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following questions:

    1)$\displaystyle \int \frac{x^2}{6x^2-7x+2}$
    This what i have done:

    do long division i get: $\displaystyle 1 - \frac{5x^2-7x+2}{6x^2-7x+2}$
    Hi

    Using long division you should find a numerator whose degree is lower than the denominator one

    $\displaystyle \frac{x^2}{6x^2-7x+2} = \frac16 + \frac{ax+b}{6x^2-7x+2}$
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following questions:

    1)$\displaystyle \int \frac{x^2}{6x^2-7x+2}$
    This what i have done:

    do long division i get: $\displaystyle 1 - \frac{5x^2-7x+2}{6x^2-7x+2}$

    $\displaystyle \int 1dx \int \frac{5x^2-7x+2}{6x^2-7x+2}$

    do partial fractions we get:

    $\displaystyle \frac{-3}{2x-1} - \frac{1}{(3x-2)}$

    final solution i got is:
    $\displaystyle x - 3ln|2x-1| - ln|3x-2|$

    2)$\displaystyle \int \frac{x-1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

    $\displaystyle u=3x^2+4x+2 du=6x+4$

    $\displaystyle \frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

    final solution i got is:
    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$

    3)$\displaystyle \int \frac{3}{(x-+1)^2)(2x-3)}$
    This what i have done:

    split into partial fractions, after i get:

    $\displaystyle A=\frac{3}{5}$ $\displaystyle B=\frac{6}{25}$ $\displaystyle C=\frac{-12}{25}
    $
    $\displaystyle \int \frac{A}{(x-1)^2} + \frac{B}{(x+1)} + \frac{C}{(2x-3)} =
    \frac{3}{5(x-1)^2} + \frac{6}{25(x+1)} + \frac{-12}{25(2x-3)}$

    final solution i got is:

    $\displaystyle \frac{3}{5(x-1)^2} + \frac{6}{25}ln|x+1| - \frac{24}{25}ln|2x-3|$

    P.S
    Your long division is off...

    $\displaystyle \frac{x^2}{6x^2 - 7x + 2} = \frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}$.


    So $\displaystyle \int{\left(\frac{x^2}{6x^2 - 7x + 2}\right)\,dx} = \int{\left[\frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}\right]\,dx}$

    $\displaystyle = \int{\left(\frac{1}{6}\right)\,dx} + \frac{1}{6}\int{\left(\frac{7x - 2}{6x^2 - 7x + 2}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7}{6}\int{\left(\frac{x - \frac{2}{7}}{6x^2 - 7x + 2}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - \frac{24}{7}}{6x^2 - 7x + 2}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - 7 + \frac{25}{7}}{6x^2 - 7x + 2}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7}{72}\int{\left(\frac{12x - 7}{6x^2 - 7x + 2}\right)\,dx} + \frac{7}{72}\int{\left(\frac{\frac{25}{7}}{6x^2 - 7x + 2}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7}{72}\int{\left(\frac{1}{u}\right)\,du} + \frac{25}{72}\int{\left(\frac{1}{6x^2 - 7x + 2}\right)\,dx}$ after making the substitution $\displaystyle u = 6x^2 - 7x + 2$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|u|}}{72} + \frac{25}{432}\int{\left(\frac{1}{x^2 - \frac{7}{6}x + \frac{1}{3}}\right)\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{x^2 - \frac{7}{6}x + \left(\frac{7}{12}\right)^2 - \left(\frac{7}{12}\right)^2 + \frac{1}{3}}\right]\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{\left(x - \frac{7}{12}\right)^2 - \frac{1}{144}}\right]\,dx}$

    Now making the substitution $\displaystyle x - \frac{7}{12} = \frac{1}{12}\cosh{t}$ so that $\displaystyle dx = \frac{1}{12}\sinh{t}\,dt$ this becomes

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{1}{\left(\frac{1}{12}\cosh{t}\right)^2 - \frac{1}{144}}\right]\,\frac{1}{12}\sinh{t}\,dt}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{144\sinh{t}}{12(\cosh^2{t} - 1)}\right]\,dt}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{432}\int{\left[\frac{12\sinh{t}}{\sinh{t}}\right]\,dt}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25}{36}\int{(1)\,dt}$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25t}{36} + C$

    $\displaystyle = \frac{x}{6} + \frac{7\ln{|6x^2 - 7x + 2|}}{72} + \frac{25\cosh^{-1}(12x - 7)}{36} + C$

    $\displaystyle = \frac{12x + 7\ln{|6x^2 - 7x + 2|} + 50\cosh^{-1}(12x - 7)}{72} + C$.
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  4. #4
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    $\displaystyle \int \frac{x^2dx}{6x^2-7z+2}$ has partial fractions
    $\displaystyle \int \frac{4}{3(3x-2)}dx - \int\frac{1}{2(2x-1)}dx + \int 1/6dx$ and thus solution:
    $\displaystyle x/6 - \frac{ln(x-1/2)}{4} + \frac{4ln(x-2/3)}{9}$.
    Note that your partial fractions can't be correct because if you try to work backwards from your result you've lost the $\displaystyle x^2$ term in the numerator
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following questions:

    1)$\displaystyle \int \frac{x^2}{6x^2-7x+2}$
    This what i have done:

    do long division i get: $\displaystyle 1 - \frac{5x^2-7x+2}{6x^2-7x+2}$
    The other posters have done a great job in completing this problem. I would just like to shed some light on long division...I absolutely, and I mean, absolutely hate it.

    So what I do is find factors such that we can reduce the numerator.

    $\displaystyle \frac{x^2}{6x^2-7x+2}$

    $\displaystyle = \frac{6}{6}[ \frac{x^2 - \frac{7}{6} x + \frac{7}{6} x + \frac{2}{6} - \frac{2}{6} }{6x^2-7x+2}]$

    $\displaystyle = \frac{1}{6} [ \frac{ 6x^2 - 7x + 2 +[7x-2] }{6x^2-7x+2}]$

    $\displaystyle = \frac{1}{6} [ 1 + \frac{ 7x -2 }{6x^2-7x+2}]$

    Which we can now decompose by partial fractions.

    No long division required
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  6. #6
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    Quote Originally Posted by GeoC View Post
    $\displaystyle \int \frac{x^2dx}{6x^2-7z+2}$ has partial fractions
    $\displaystyle \int \frac{4}{3(3x-2)}dx - \int\frac{1}{2(2x-1)}dx + \int 1/6dx$ and thus solution:
    $\displaystyle x/6 - \frac{ln(x-1/2)}{4} + \frac{4ln(x-2/3)}{9}$.
    how did you get this solution i always get $\displaystyle \frac{-3}{2x-1} + \frac{8}{3x-2} $from doing partial fractions.
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    how did you get this solution i always get $\displaystyle \frac{-3}{2x-1} + \frac{8}{3x-2} $from doing partial fractions.
    $\displaystyle \frac{x^2}{6x^2 - 7x + 2} = \frac{1}{6} + \frac{7x - 2}{6(6x^2 - 7x + 2)}$

    $\displaystyle = \frac{1}{6} + \frac{1}{6}\left[\frac{7x - 2}{(3x- 2)(2x - 1)}\right]$


    Now using partial fractions:

    $\displaystyle \frac{A}{3x - 2} + \frac{B}{2x - 1} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

    $\displaystyle \frac{A(2x - 1) + B(3x - 2)}{(3x - 2)(2x - 1)} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

    $\displaystyle A(2x - 1) + B(3x - 2) = 7x - 2$

    $\displaystyle 2Ax - A + 3Bx - 2B = 7x - 2$

    $\displaystyle (2A + 3B)x - (A + 2B) = 7x - 2$.


    So now your system of equations is

    $\displaystyle 2A + 3B = 7$

    $\displaystyle A + 2B = 2$.


    Multiply the second equation by 2...


    $\displaystyle 2A + 3B = 7$

    $\displaystyle 2A + 4B = 4$.


    Subtract the second equation from the first...


    $\displaystyle (2A + 3B) - (2A + 4B) = 7 - 4$

    $\displaystyle -B = 3$

    $\displaystyle B = -3$.


    Substituting back into equation 2...

    $\displaystyle A + 2B = 2$

    $\displaystyle A + 2(-3) = 2$

    $\displaystyle A - 6 = 2$

    $\displaystyle A = 8$.



    So $\displaystyle \frac{7x - 2}{(3x - 2)(2x - 1)} = \frac{8}{3x - 2} - \frac{3}{2x - 1}$

    and therefore

    $\displaystyle \frac{1}{6} + \frac{1}{6}\left[\frac{7x - 2}{(3x- 2)(2x - 1)}\right] = \frac{1}{6} + \frac{1}{6}\left(\frac{8}{3x - 2} - \frac{3}{2x - 1}\right)$

    $\displaystyle = \frac{1}{6} + \frac{4}{3(3x - 2)} - \frac{1}{2(2x - 1)}$.



    So that means

    $\displaystyle \int{\frac{x^2}{6x^2 - 7x + 2}\,dx} = \int{\frac{1}{6}\,dx} + \frac{4}{3}\int{\frac{1}{3x - 2}\,dx} - \frac{1}{2}\int{\frac{1}{2x - 1}\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{4}{9}\int{\frac{3}{3x - 2}\,dx} - \frac{1}{4}\int{\frac{2}{2x - 1}\,dx}$

    $\displaystyle = \frac{x}{6} + \frac{4\ln{|3x - 2|}}{9} - \frac{\ln{|2x - 1|}}{4} + C$

    $\displaystyle = \frac{6x + 16\ln{|3x - 2|} - 9\ln{|2x - 1|}}{36} + C$.
    Last edited by Prove It; May 29th 2010 at 08:49 PM.
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  8. #8
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    Subtract the second equation from the first...


    $\displaystyle (2A + 3B) - (2A + 4B) = 7 - 4$

    $\displaystyle -B = 3$

    $\displaystyle B = 3$.


    is this meant to be -3 not 3
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  9. #9
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    this is how i did it, i don't understand how my way is incorrect.

    $\displaystyle \frac{A}{2x-1} + \frac{B}{3x-2}$

    $\displaystyle A(3x-2) + B(2x-1) = 7x-2$

    3A + 2B = 7 (1
    -2A - B = -2 (2

    times 2

    -4A - B = -4

    minus 1) from 2)

    -A = 3
    A = -3

    therefore B = $\displaystyle \frac{16}{2} = 8$

    so i get

    $\displaystyle \frac{-3}{(2x-1)} + \frac{8}{(3x-2)}$
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  10. #10
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    Quote Originally Posted by Paymemoney View Post
    this is how i did it, i don't understand how my way is incorrect.

    $\displaystyle \frac{A}{2x-1} + \frac{B}{3x-2}$

    $\displaystyle A(3x-2) + B(2x-1) = 7x-2$

    3A + 2B = 7 (1
    -2A - B = -2 (2

    times 2

    -4A - B = -4

    minus 1) from 2)

    -A = 3
    A = -3

    therefore B = \frac{16}{2} = 8

    so i get

    $\displaystyle \frac{-3}{(2x-1)} + \frac{8}{(3x-2)}$
    Your algebra is not correct. You are messing up your signs.
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    is this meant to be -3 not 3
    Yes it is - edited.
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  12. #12
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    Quote Originally Posted by Paymemoney View Post
    this is how i did it, i don't understand how my way is incorrect.

    $\displaystyle \frac{A}{2x-1} + \frac{B}{3x-2}$

    $\displaystyle A(3x-2) + B(2x-1) = 7x-2$

    3A + 2B = 7 (1
    -2A - B = -2 (2

    times 2

    -4A - 2B = -4

    minus 1) from 2) Since the signs are different for B here, you need to add the equations.
    From here you should have gotten:

    $\displaystyle -A = 3$ which means $\displaystyle A = -3$.

    Sub into 2)

    $\displaystyle -2(-3) - B = -2$

    $\displaystyle 6 - B = -2$

    $\displaystyle B = 8$.
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  13. #13
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    Just to give a different method

    From

    $\displaystyle \frac{A}{3x - 2} + \frac{B}{2x - 1} = \frac{7x - 2}{(3x - 2)(2x - 1)}$

    I multiply both sides by $\displaystyle 3x - 2$

    $\displaystyle A + \frac{B(3x-2)}{2x - 1} = \frac{7x - 2}{2x - 1}$

    And I set $\displaystyle x = \frac23$

    $\displaystyle A = \frac{7 \frac23 - 2}{2 \frac23 - 1} = \frac{\frac83}{\frac13} = 8$

    I do the same for $\displaystyle 2x - 1$

    $\displaystyle \frac{A(2x-1)}{3x - 2} + B = \frac{7x - 2}{3x - 2}$

    And I set $\displaystyle x = \frac12$

    $\displaystyle B = \frac{7 \frac12 - 2}{3 \frac12 - 2} = \frac{\frac32}{-\frac12} = -3$
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  14. #14
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    Can anyone answer my other two question??
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  15. #15
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    Quote Originally Posted by Paymemoney View Post
    2)$\displaystyle \int \frac{x-1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6} \int \frac{6x+4}{3x^2+4x-2} + \int \frac{1}{3x^2+4x-2}$

    $\displaystyle u=3x^2+4x+2 du=6x+4$

    $\displaystyle \frac{1}{6} \int \frac{du}{u} + \int \frac{1}{3x^2+4x+2}$

    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \int \frac{1}{(3x+\frac{2}{\sqrt{3}}+\frac{14}{9})}$

    final solution i got is:
    $\displaystyle \frac{1}{6}ln|3x^2+4x+2| + \frac{1}{9}arcttan(3x+ \frac{2}{\sqrt{3}}) +c$
    $\displaystyle \int {\frac{x-1}{3x^2+4x+2}\,dx} = \frac{1}{6}\int{\frac{6x - 6}{3x^2 + 4x + 2}\,dx}$

    $\displaystyle = \frac{1}{6}\int{\frac{6x + 4}{3x^2 + 4x + 2}\,dx} - \frac{1}{6}\int{\frac{10}{3x^2 + 4x + 2}\,dx}$

    $\displaystyle = \frac{1}{6}\int{\frac{1}{u}\,du} - \frac{5}{9}\int{\frac{1}{x^2 + \frac{4}{3}x + \frac{2}{3}}\,dx}$ after making the substitution $\displaystyle u = 3x^2 + 4x + 2$

    $\displaystyle = \frac{\ln|u|}{6} - \frac{5}{9}\int{\frac{1}{x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + 2}\,dx}$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{1}{\left(x + \frac{2}{3}\right)^2 + \frac{14}{9}}\,dx}$

    Now make the substitution $\displaystyle x + \frac{2}{3} = \frac{\sqrt{14}}{3}\tan{\theta}$ so that $\displaystyle dx = \frac{\sqrt{14}}{3}\sec^2{\theta}\,d\theta$ and the integral becomes

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{1}{\frac{14}{9}\tan^2{\theta } + \frac{14}{9}}\,\frac{\sqrt{14}}{9}\sec^2{\theta}\, d\theta}$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5}{9}\int{\frac{\sqrt{14}\sec^2{\theta}}{14( \tan^2{\theta} + 1)}\,d\theta}$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}}{126}\int{\frac{\sec^2{\theta}}{\ sec^2{\theta}}\,d\theta}$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}}{126}\int{1\,d\theta}$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}\,\theta}{126} + C$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\sqrt{14}\left[\frac{3}{\sqrt{14}}\arctan{\left(x + \frac{2}{3}\right)}\right]}{126} + C$

    $\displaystyle = \frac{\ln{|3x^2 + 4x + 2|}}{6} - \frac{5\arctan{\left(x + \frac{2}{3}\right)}}{42} + C$

    $\displaystyle = \frac{7\ln{|3x^2 + 4x + 2|} - 5\arctan{\left(x + \frac{2}{3}\right)}}{42} + C$.
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