Understanding Partial Fraction Decomposition in Integrals

[stripedcat]

First the example problem. This is an integral of the whole thing

(3x^3+24x^2+56x-5) / (x^2+8x+17)^2

The answer comes out to be

3/2 ln(x^2+8x+17) - (49/2 tan^-1(x+4)) - (25x+105 / 2(x^2+8x+17) + C

I would show all the steps but I'm still not sure on how to use the format tools, so that would get really messy to read.

The 'real' problem is

(3x^3+18x^2+37-4) / (x^2+6x+10)^2

Which I solve to be...

3/2 ln(x^2+6x+10)- (43/2 tan^-1(x+3)) - (25x+82 / 2(x^2+6x+10)) + C

Which is 'wrong', I keep coming back with that same answer though, and online resources seem to confirm it?

I just don't know where I'm going wrong.

 

[MarkFL]

We can't really tell you where you are going wrong without seeing your work. :D

The first step is to obtain the partial fraction decomposition of the integrand...what did you get for that?

 

[stripedcat]

We can't really tell you where you are going wrong without seeing your work. :D

The first step is to obtain the partial fraction decomposition of the integrand...what did you get for that?

For the second problem?

(3x^3+18x^2+37-4) / (x^2+6x+10)^2 =

((3x / x^2+10x) / (x^2+6x+10)) + ((7x-4) / (x^2+6x+10)^2)

That is what you wanted?

 

[MarkFL]

Okay, the actual integrand must be:

\(\displaystyle \frac{3x^3+18x^2+37x-4}{\left(x^2+6x+10\right)^2}\)

You have the second term correct, but the actual decomposition is:

\(\displaystyle \frac{3x}{x^2+6x+10}+\frac{7x-4}{\left(x^2+6x+10\right)^2}\)

Now, I would suggest adding zero to each numerator in a form that makes integrating easier...what do you get after doing this?

 

[stripedcat]

Scratch all that noise.

My answer was 'wrong', and yes, there was a uh... 'not so kind' email over this.

Typing out tan^-1 vs using the system to make a 'tan^-1( )

You tell me if you can tell the difference, maybe I just don't see it.

View attachment 2791

 

[Prove It]

Scratch all that noise.

My answer was 'wrong', and yes, there was a uh... 'not so kind' email over this.

Typing out tan^-1 vs using the system to make a 'tan^-1( )

You tell me if you can tell the difference, maybe I just don't see it.

View attachment 2791

For the inverse tangent function, try Arctan(x). Your input is probably making the computer read it as [tan(x)]^(-1).

 

[stripedcat]

Anything is possible.

Had a problem last week with other trig related inputs.

The examples always showed 'sinx', but if you put in 'sinx' for your answer, it counted it wrong... But only sometimes.

But it always counted sin(x) right.

 

[MarkFL]

Look at the last denominator in the incorrect submission...there is a missing $x$. :D