# Thread: Integration of Rational Functions By Partial Fractions

1. ## Integration of Rational Functions By Partial Fractions

I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

"Find $\int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $x$, use the constant $C$."

So, I have:

$
\int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
$

and

$
6x+1 = A(4x-2) + B(5x+6)
$

If $x=\frac{1}{2}$, then $B = \frac{8}{17}$

and

If $x = \frac{-6}{5}$, then $A = \frac{31}{850}$

So,

$
\int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

$
= \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
$

Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

Austin Martin

2. Originally Posted by auslmar
I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

"Find $\int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $x$, use the constant $C$."

So, I have:

$
\int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
$

and

$
6x+1 = A(4x-2) + B(5x+6)
$

If $x=\frac{1}{2}$, then $B = \frac{8}{17}$

and

If $x = \frac{-6}{5}$, then $A = \frac{31}{850}$

So,

$
\int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

$
= \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
$

Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

Austin Martin
B is OK. A is wrong .... A = 31/34.

3. Originally Posted by mr fantastic
B is correct. A is wrong .... A = 31/34.

${\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}8}}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}16}}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$

where you keep re-arranging until you recognise the correct answer.

Note the use of absolute value!

4. Originally Posted by auslmar
I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

"Find $\int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $x$, use the constant $C$."

So, I have:

$
\int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
$

and

$
6x+1 = A(4x-2) + B(5x+6)
$

If $x=\frac{1}{2}$, then $B = \frac{8}{17}$

and

If $x = \frac{-6}{5}$, then $A = \frac{31}{850}$

So,

$
\int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

$
= \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
$

Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

Austin Martin
If you need to I will go through the entire PFD for you, but a first thing that will make it a little bit easier. Why not finish your factorization?

$4x-2=2(2x-1)$

So you can forego it

5. Thanks to both of you. Now that it's established that I suck at arithmetic, I'm having trouble seeing why the answer is not:

$
\frac{31}{34}ln|5x+6| + \frac{8}{17}ln|4x-2| + C
$
since $A = \frac{31}{34}$ and $B = \frac{8}{17}$
Also Mr. Fantastic, I'm not sure what you mean by:

Originally Posted by mr fantastic

$\frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + \frac{4}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + \frac{8}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$

where you keep re-arranging until you recognise the correct answer.

Note the use of absolute value!
Re-arranging? Forgive my dense-ness.

6. Originally Posted by auslmar
Thanks to both of you. Now that it's established that I suck at arithmetic, I'm having trouble seeing why the answer is not:

$
\frac{31}{34}ln|5x+6| + \frac{8}{17}ln|4x-2| + C
$
since $A = \frac{31}{34}$ and $B = \frac{8}{17}$

Mr F says: That's one form of the correct answer.

Also Mr. Fantastic, I'm not sure what you mean by:

Re-arranging? Forgive my dense-ness.
Other forms of the correct answer can be found by re-arranging the above. For example:

$\frac{1}{34} \left( 31 \ln|5x+6| + 16 \ln|4x-2| + C \right)$.

If you state what form the final answer is given in, you can be shown how to get that form from the above.

7. Originally Posted by mr fantastic
Other forms of the correct answer can be found by re-arranging the above. For example:

$\frac{1}{34} \left( 31 \ln|5x+6| + 16 \ln|4x-2| + C \right)$.

If you state what form the final answer is given in, you can be shown how to get that form from the above.
Oh, I see. I use a homework system called Maple TA that allows the user to preview if their answer is correct. Unfortunately, when I enter any forms of the previously determined correct answer, the system says they're wrong. I can't figure out what's wrong. I'm using absolute values and including the costant $C$, so I'm puzzled as to why it's not recognizing the answer.

8. Originally Posted by mr fantastic

${\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{4}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}16}}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$

where you keep re-arranging until you recognise the correct answer.

Note the use of absolute value!
My mistake. See the edit above. I was very careless. So were you, of course.

In fact, if you follow mathstud's suggestion of simplifying 4x - 2 = 2(x - 1), you will probably get the form of the answer that's being looked for a lot easier ......

Note that $\frac{1}{4} \, \frac{16}{34} \, \ln | 4x - 2 | + C = \frac{4}{34} \, \ln 2| 2x - 1 | + C$

$= \frac{4}{34} \, \ln| 2x - 1 | + \frac{4}{34} \, \ln (2) + C$ where the last is just a constant, D say

$= \frac{4}{34} \, \ln| 2x - 1 | + D = \frac{20}{170} \, \ln| 2x - 1 | + D$

That should give you blue sky .........