Question

Hello, can anyone tell me how to integrate 3x+8/(x^2+5x+6)??? I tried using substitution but got confused

Answer 1

split it into partial fractions.

Answer 2

\[\int{\frac{1}{x+3}+\frac{2}{x+2}}dx\]

Answer 3

I dont quite understand what happens to the 3x+8?

Answer 4

It is split into partial fractions.

Answer 5

The whole fraction, I mean.

Answer 6

x^2 + 5x +6 = x^2 + 3x +2x +6 = x(x+3) + 2(x+3) = (x+2)(x+3)
now you can turn it into two fractions :
A/(x+2) + B/ (x+3) = (3x+8) / (x+3)(x+2)
A(x+3) + B(x+2) = 3x + 8
Ax + 3A + Bx + 2B = 3x + 8
A+B = 3 -> B = 3-A
3A+2B = 8
3A + 2(3-A) = 8
3A + 6 -2A = 8
A = 2
B = 1
2/(x+2) + 1/(x+3)

Answer 7

ok I see, thank you very much. Does partial fractions apply to most of these fraction problems?

Answer 8

Polynomial fractions, maybe.

Answer 9

Now, the reason for my questions was to solve a problem involving differential equations and can't seem to get the right answer: The problem is to solve the diff. eq.:
(3x+8)(y^2+4)dx - 4y(x^2+5x+6)dy = 0 with initial condition y(1)=2
I believe this equation is separable and by using an integrating factor I got
[3x+8/(x+2)(x+3)]dx = [4y/(y^2+4)] dy. Integrating this I got
(x+2)^2 + (x+3) +C = (y^2 + 4)^2 and I suppose the given initial condition is to find the value of C but even though I use it does not have nothing to do with the answer in the book which is
16(x+3)(x+2)^2 = 9(y^2 + 4)^2. Where does the 16 and 9 from each side respectively come from?

Answer 10

I got it .... thanks!

Answer 11

integrating :
2ln(x+2) + ln(x+3) +C = 2ln(y^2+4)
2ln(x+2) + ln(x+3) +lnC = 2ln(y^2+4)
(x+2)^2(x+3)C=(y^2+4)
x =1 , y = 2
9*4*C = 64
C = 16/9

Answer 12

ty!
I am really having problems integrating.. how do you integrate y/(y+2) ??? I tried integrating by parts but did not work

Answer 13

y/(y+2) = 1- 2/(y+2)

Answer 14

I see , again partial fractions. I just wasnt sure since there was only a y+2 at the bottom. Thank you , you are the best!

Answer 15

you are welcome :)