# 3 Questions

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• Sep 10th 2009, 05:06 PM
arze
3 Questions
Use partial fractions to find:
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx$
$\displaystyle \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}$
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$
$\displaystyle =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid$
$\displaystyle =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid$
Answer is supposed to be $\displaystyle 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid$

$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx$
$\displaystyle \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$
$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx$
$\displaystyle =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid$
$\displaystyle =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid$
Answer is supposed to be $\displaystyle \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid$

$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$\displaystyle =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $\displaystyle 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
Thanks
• Sep 10th 2009, 05:26 PM
Prove It
Quote:

Originally Posted by arze
Use partial fractions to find:
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx$
$\displaystyle \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}$
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$
$\displaystyle =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid$
$\displaystyle =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid$
Answer is supposed to be $\displaystyle 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid$

$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx$
$\displaystyle \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$
$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx$
$\displaystyle =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid$
$\displaystyle =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid$
Answer is supposed to be $\displaystyle \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid$

$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$\displaystyle =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $\displaystyle 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
Thanks

In the first one, everything is right up to this step...

$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$

I don't know where you got the $\displaystyle 13\ln{|3-x|}$ part...

It should be

$\displaystyle = -12\ln{|2 - x|} + 36\ln{|3 - x|} - 24\ln{|4 - x|} + C$

$\displaystyle = 12\left[-\ln{|2 - x|} + 3\ln{|3 - x|} - 2\ln{|4 - x|}\right] + C$

$\displaystyle = 12\left[-\ln{|2 - x|} + \ln{|3 - x|^3} - \ln{|4 - x|^2}\right] + C$

$\displaystyle = 12\left[\ln{\frac{|3 - x|^3}{|2 - x||4 - x|^2}}\right] + C$

$\displaystyle = 12\ln{\left|\frac{(3 - x)^3}{(2 - x)(4 - x)^2}\right|} +C$.
• Sep 10th 2009, 05:33 PM
Prove It
Quote:

Originally Posted by arze
Use partial fractions to find:
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx$
$\displaystyle \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}$
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$
$\displaystyle =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid$
$\displaystyle =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid$
Answer is supposed to be $\displaystyle 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid$

$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx$
$\displaystyle \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$
$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx$
$\displaystyle =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid$
$\displaystyle =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid$
Answer is supposed to be $\displaystyle \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid$

$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$\displaystyle =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $\displaystyle 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
Thanks

For Q.2, Again, correct up to

$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx$

This should be

$\displaystyle = -\frac{7}{2}\ln{|2x - 1|} + \frac{7}{2}\ln{|x - 1|} + \frac{1}{2}\ln{|x + 1|} + C$

$\displaystyle = \frac{1}{2}\left[-7\ln{|2x - 1|} + 7\ln{|x - 1|} + \ln{|x + 1|}\right] + C$

$\displaystyle = \frac{1}{2}\left[-\ln{|2x - 1|^7} + \ln{|x - 1|^7} + \ln{|x + 1|}\right] + C$

$\displaystyle = \frac{1}{2}\left[\ln{\left(\frac{|x - 1|^7|x + 1|}{|2x - 1|^7}\right)}\right]+C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(x -1)^7(x + 1)}{(2x - 1)^7}\right|} + C$.
• Sep 10th 2009, 05:35 PM
seld
for your first one at the very least, your fractions are wrong.
I'm pretty sure it's supposed to be 12/(2-x)+-36/(3-x)+24/(4-x)

going to take a bit to check the others.
• Sep 10th 2009, 05:37 PM
Prove It
Quote:

Originally Posted by arze
Use partial fractions to find:
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx$
$\displaystyle \frac{12x}{(2-x)(3-x)(4-x)}=\frac{12}{2-x}+\frac{-36}{3-x}+\frac{24}{4-x}$
$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$
$\displaystyle =12 \ln \mid 2-x\mid-13\ln \mid 3-x\mid+24\ln \mid 4-x\mid$
$\displaystyle =12\ln \mid \frac{(2-x)(4-x)^2}{(3-x)^3}\mid$
Answer is supposed to be $\displaystyle 12\ln \mid \frac{(3-x)^3}{(2-x)(4-x)^2}\mid$

$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx$
$\displaystyle \frac{x^2+2x+4}{(2x-1)(x^2-1)}\equiv \frac{-7}{2x-1}+\frac{\frac{7}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$
$\displaystyle \int \frac{x^2+2x+4}{(2x-1)(x^2-1)} dx\equiv -\int \frac{7}{2x-1}dx+\int \frac{\frac{7}{2}}{x-1}dx+\int \frac{\frac{1}{2}}{x+1}dx$
$\displaystyle =-7\ln \mid 2x-1\mid +\frac{7}{2}\ln \mid x-1\mid+\frac{1}{2}\ln \mid x+1\mid$
$\displaystyle =\ln \mid \frac{(x-1)^{\frac{7}{2}}(x+1)^{\frac{1}{2}}}{(2x-1)^7}\mid$
Answer is supposed to be $\displaystyle \frac{1}{2}\ln \mid \frac{(x-1)^7(x+1)}{(2x-1)^7}\mid$

$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\displaystyle \frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\displaystyle \int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$\displaystyle =2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $\displaystyle 2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
I've worked through them several times and always get the same answer, but its not the answer in the book. Can anyone help me see if I've made any errors?
Thanks

For Q.3, you have done everything right except the final integration.

You made the same mistake in ALL of the questions.

You need to remember that

$\displaystyle \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\ln{|ax + b|} + C$.

You keep forgetting to take out the common factor of $\displaystyle \frac{1}{a}$...
• Sep 10th 2009, 06:35 PM
arze
Quote:

Originally Posted by Prove It
In the first one, everything is right up to this step...

$\displaystyle \int \frac{12x}{(2-x)(3-x)(4-x)} dx\equiv \int \frac{12}{2-x}dx-\int \frac{36}{3-x}dx+\int \frac{24}{4-x}dx$

I don't know where you got the $\displaystyle 13\ln{|3-x|}$ part...

It should be

$\displaystyle = -12\ln{|2 - x|} + 36\ln{|3 - x|} - 24\ln{|4 - x|} + C$

$\displaystyle = 12\left[-\ln{|2 - x|} + 3\ln{|3 - x|} - 2\ln{|4 - x|}\right] + C$

$\displaystyle = 12\left[-\ln{|2 - x|} + \ln{|3 - x|^3} - \ln{|4 - x|^2}\right] + C$

$\displaystyle = 12\left[\ln{\frac{|3 - x|^3}{|2 - x||4 - x|^2}}\right] + C$

$\displaystyle = 12\ln{\left|\frac{(3 - x)^3}{(2 - x)(4 - x)^2}\right|} +C$.

The 13 was a typo. Thanks!
• Sep 10th 2009, 06:36 PM
arze
Now I see where I got it wrong. Thanks much :D