Integration Question

Dec 2008
509
2
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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Nov 2008
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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

\(\displaystyle \frac{x^2}{6x^2-7x+2} = \frac16 + \frac{ax+b}{6x^2-7x+2}\)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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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May 2010
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1
\(\displaystyle \int \frac{x^2dx}{6x^2-7z+2}\) 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
 
Apr 2010
384
153
Canada
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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Dec 2008
509
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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:
\(\displaystyle x/6 - \frac{ln(x-1/2)}{4} + \frac{4ln(x-2/3)}{9}\).
how did you get this solution i always get \(\displaystyle \frac{-3}{2x-1} + \frac{8}{3x-2} \)from doing partial fractions.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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:
Dec 2008
509
2
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
 
Dec 2008
509
2
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

\(\displaystyle \frac{-3}{(2x-1)} + \frac{8}{(3x-2)}\)
 
Apr 2010
384
153
Canada
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.