Integration by Partial Fractions

[Illusionist]

Homework Statement

[(3x^2)+10x+13]/[(x-1)([x^2]+4x+8)]

Homework Equations

I think solving this question should include partial fractions.

The Attempt at a Solution

I've made a few different attempts at this question but find myself at a dead end every time.
One attempt was expanding the denominator (giving me [x^3]+3[x^2]+4x-8. I then let this equal v, hence dv/du= 3(u^2)+6u+4.
From this I got my original equation to look like this: [((3x^2)+6x+4)+(4x+9)]/[(x^3)+3(x^2)+4x-8] hence:

1/v .dv + (4x+9)]/[(x^3)+3(x^2)+4x-8] . dx

I can seem to differentiate the second part, but even worse I think this approach is wrong.

If anyone has any advice or help with this sort of question please share, I would greatly appreciate it. Thanks guys.

 

[Winzer]

Do you mean:
$$\int \frac{3x^2+10x+13}{(x-1)x^2+4x+8}dx$$?

 

[matness]

I think solving this question should include partial fractions

Right.
remember how to do partial fraction
$$\frac{3x^2+10x+13}{(x-1)(x^2+4x+8)} = \frac{A}{x-1}+\frac{Bx+C}{x^2+4x+8}$$
Find A, B and C and try to solve then we can continue.

 

[Winzer]

Matness is correct. You will have that linear term and the quadratic term.
Get your common denominator and cancell with the left original equation.
And in order to find A & B, the two polynomials inorder to be equal must have equal coefficients for each power of the polynomial..I assume you can take it from here?

 

[Illusionist]

Right.
remember how to do partial fraction
$$\frac{3x^2+10x+13}{(x-1)(x^2+4x+8)} = \frac{A}{x-1}+\frac{Bx+C}{x^2+4x+8}$$
Find A, B and C and try to solve then we can continue.

Alright here is what I have come up with:

(3x^2 + 10x + 13) = A(x^2 + 4x + 8) + (Bx + C)(x-1)
(3x^2 + 10x + 13) = A(x^2) + 4xA + 8A + B(x^2) - Bx + Cx - C

Hence - (x^2): 3 = A + B -------> B = 3 - A
(x^1): 10 = 4A - B + C
(x^0): 13 = 8A - C -------> C = 13 - 8A

Sub'ing back in I get:
10 = 4A - (3-A) + (13-8A)
10 = -3A + 10
A= 0
Therefore B=3-0=3 and C = 13-0=13

Hence now we are looking for the integral of (3x + 13)/(x^2 + 4x + 8)

Now this is where I'm running into problems, here's what I've done:
let v= x^2 + 4x + 8 hence dv/du= 2x+4

Hence 2/3 ( [2x + (39/x)] / v ) . du = 2/3 ( [ 2x + 4/ v ] + 9 ( 2^2 + (u+2)^2 ) ) . du

Which after integration I ended up with (2/3) log(v) + 3arctan ([x+2]/2) + C.

This isn't the answer I'm looking for and not sure where I went wrong. Sorry about the long working guys. Again help would be really great and thanks for the responses already guys.

 

[matness]

I think your solution is nice except one point

13 = 8A - C -------> C = 13 - 8A

If you correct as C=8A-13
You possibly find what you expectd
And thank you for showing your work
That is what we want to see here usually

edit: you have actually some other arithmetical mistakes in the second part
Anyway forget them and try to solve with your new coefficients

 

[Illusionist]

OK yeah C = 8A - 13 I recalculated:
10= 4Ax - (3 - A)x + x(8A-13)
10= 3Ax 16x
A = (16/13) + (10/13)x

This value of x gave me the following for B and C:
B = 3 - [(16/13) + (10/13)x] = (23/13) - (10/13)x

C = 8[(16/13) + (10/13)x] - 13
C = (80/13)x - (41/13)

Now I still think something is horribly wrong because these sort of solutions for A, B and C are very ugly and makes thinks terrible hard and ugly.

 

[matness]

where did you find x's?

I think we were agree that
-------------
A+B=3------>B=3-A
4A - B + C =10
8A - C =13--->C=8A-13
----------------

4A-(3-A)+8A-13=10 ---> 13A=26 -->A=2, B=1, C=3

Now try to integrate but not hurry in order to obtain a correct result