# Partial Fractions Webwork Problem 3

Hey everyone.  This is Andrew Maloney posting with a question:

I can’t figure out what to do with Partial Fractions question 3.  I know that my denominator has no Real roots So it’s a product of one irreducible (over the Reals) quadratic.  It seems to me that the partial fractions technique won’t get me anywhere with this problem.

I think I should parse the denominator as:

(x+3)^2 + 25

Then use substitution and trig substitution.

Yes, doing this gives me the answer.

Can anyone see a way that partial fractions could be applied here?  In fact, as I move through the other questions, I also can’t find a way to apply the partial fractions technique to question 10. Instead, I use a substitution within a substitution, similar to the technique I used for this problem but with regular substitution instead of trig substitution.

What am I missing?

Thanks!

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### 8 Responses to Partial Fractions Webwork Problem 3

1. avi says:

Hi i am having a problem with number 6) i get the answer as: ((1/13)/(x+5))+((1/-3)/(x+8))

• Maloney says:

Avi, you need to let us know what problem you received for #6. We each get slightly different versions. Perhaps it would be better to create a second post for 6. I don’t want things to get drawn away from question 3. Thanks!

• Maloney says:

Avi, a couple thoughts:

You might need those coefficients up front. (1/13)ln…

Definitely the answer will have the natural log twice. Your first answer doesn’t appear to have ln at all.

Your second formulation doesn’t have the power carrot ^, which I guess turns those coefficients into exponents.

Try pressing preview, and check that your answer is being interpreted the way you intend.

Also, and most important, are you entering the most general solution? (+ C)

I hope this helps!

• coolal says:

problem number 6) 1/((x+5)(x+8))

• Maloney says:

What do you get for A and B?

2. Kate Poirier says:

Andrew, it sounds like you’re on the right track. The method of partial fractions applies only when the integrand is a rational function AND the denominator actually factors. This means that if you try setting up a partial fraction decomposition like we did in class for the above question, you’ll be writing $latex \frac{1}{x^2+6x+34} = \frac{Ax+B}{x^2+6x+34}$ so $latex A=0$ and $latex B=1$. That’s not a lot of help; it just says $latex \frac{1}{x^2+6x+34} = \frac{1}{x^2+6x+34}$. For integrals like these, you have to resort to another method. Since there’s no factor of $latex x$ in the numerator here, it looks like you’ll have to complete the square in the denominator and try trigonometric substitution.

I’m glad that WebWork gave you a question like this because it forces you to remember all the techniques you learned previously.

3. Kate Poirier says:

Hrm, it seems my LaTeX is not displaying correctly. Hopefully it’s obvious enough what I mean! I’ll see if I can figure out which setting I can change to fix it…

• Maloney says:

Thank you!