3 Questions

**arze** · September 10th 2009, 05:06 PM

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

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

$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$=2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
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

**Prove It** · September 10th 2009, 05:26 PM

Originally Posted by arze

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

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

$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$=2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
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...

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

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

It should be

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

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

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

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

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

**Prove It** · September 10th 2009, 05:33 PM

Originally Posted by arze

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

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

$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$=2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
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

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

This should be

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

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

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

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

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

**seld** · September 10th 2009, 05:35 PM

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.

**Prove It** · September 10th 2009, 05:37 PM

Originally Posted by arze

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

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

$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du$
$\frac{4u^2+3u-2}{(u+1)(2u+3)}=\frac{4u^2+3u-2}{2u^2+5u+3}=2+\frac{-7u-8}{2u^2+5u+3}=2+\frac{-1}{u+1}+\frac{-5}{2u+3}$
$\int \frac{4u^2+3u-2}{(u+1)(2u+3)} du=\int 2du-\int \frac{1}{u+1}du-5\int \frac{1}{2u+3}du$
$=2u-\ln \mid u+1\mid-5\ln \mid 2u+3\mid$
Answer is supposed to be $2u-\ln \mid u+1\mid-\frac{5}{2} \ln \mid 2u+3\mid$
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

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

You keep forgetting to take out the common factor of $\frac{1}{a}$ ...

**arze** · September 10th 2009, 06:35 PM

Originally Posted by Prove It

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

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

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

It should be

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

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

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

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

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

The 13 was a typo. Thanks!

**arze** · September 10th 2009, 06:36 PM

Now I see where I got it wrong. Thanks much

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