# Calc II Partial Fractions

1. Sep 13, 2011

### PCSL

$\int\frac{7x^2+4x+5}{x^3+x} dx$

I have to solve this using partial fractions and all I could think of was breaking it up into:

$\frac{A}{x}+\frac{B}{x^2+1}$

also I tried

$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x}$
The reasoning behind this one was to split up original fraction into $\frac{7x^2+4x+5}{(x^3+0)+x} dx$

Are either of these two even right? and if so I have no clue how to proceed.

Forgive me for not showing more work, but I have been staring at this for 30 minutes and can think of nothing. Thank you.

2. Sep 13, 2011

### Staff: Mentor

For irreducible quadratic factors in the denominator (like x2 + 1), the numerator should be Ax + B. IOW, your decomposition should be:
$$\frac{A}{x} + \frac{Bx + C}{x^2+1}$$

3. Sep 13, 2011

### micromass

Staff Emeritus
Try breaking it up into

$$\frac{A}{x}+\frac{Bx+C}{x^2+1}$$

4. Sep 13, 2011

### PCSL

Alright so then if I multiplied through by x would i have:

$$\frac{7x^2+4x+5}{x^2+1}=A + \frac{x(Bx + C)}{x^2+1}$$

5. Sep 13, 2011

### Staff: Mentor

You need to solve for A, B, and C, and these values do not depend on x. Just do the algebra first.
$$\frac{7x^2+4x+5}{x^3+x} =\frac{A}{x} + \frac{Bx + C}{x^2+1}$$

You're solving for A, B, and C so that this equation is an identity - true for all values of x other than those that make the denominators zero.

The first thing to do is to multiply both sides by x(x2 + 1)

6. Sep 13, 2011

### PCSL

When I solve for A, B and C - I get A=5, B=7, and C=4.

For the final answer i get:

$$5ln(x)+4arctan(x)+\frac{7}{2}ln(x^2+1)$$ when the real answer is $$5ln(x)+4arctan(x)+ln(x^2+1)$$ Thanks for helping me get almost to the end but can you help me figure out why my answer is different? I solved for B by dividing both sides by $$x^2$$

7. Sep 13, 2011

### PCSL

Not to be a nuisance but I still need help.

8. Sep 13, 2011

### Rayquesto

Yea this stuff just gets a little getting used to. What do you need help with still? Are you unclear of the decomposing fractions?

9. Sep 13, 2011

### PCSL

Thanks, this is what I don't understand:

10. Sep 13, 2011

### Rayquesto

I think you screwd up a bit on the algebra.
A=5
A + B=7
C=4

11. Sep 13, 2011

### PCSL

Alright, is the equation

$$7x^2+4x+5=A(x^2+1)+Bx^2+Cx$$ correct for when I multiply both sides by $$x(x^2+1)$$ I'm kinda lost here for how I would solve for B. Wouldn't I divide by x^2, then set x=0, and that would give B=7?

12. Sep 13, 2011

### Rayquesto

yes.

if 7x^2 + 4x +5 = Ax^2 + A + Bx^2 +Cx

then how do you find the coefficients of A, B and C? (hint: the coefficients are directly associated with the powers of exponents).

13. Sep 13, 2011

### PCSL

I get C=4 and A=5 easy. I can not for the life of be understand why B does not equal 7. I have no clue what your hint means. Why can't you divide both sides by x^2 and get B=7 if you set x=0??

14. Sep 13, 2011

### Staff: Mentor

First off, if you set x = 0, you can't then come around and divide by x2.
The equation you are solving is
7x2 + 4x + 5 = (A + B)x2 + Cx + A
This has to be identically true, so the coefficients of x2 and x and the constants have to be equal.

15. Sep 13, 2011

### PCSL

Thank you for you help, I think I have a better understanding of solving when the denominator is irreducible but my real problem was that the denominator in this particular problem was x^3-x not x^3+x..... thank you all for your help, however.