Question

Help with two partial fractions problems please :)

Answer 1

1) \[\int\limits_{1}^{2} \frac{ (x+1) }{ x(x^2 + 1) } dx\]

Answer 2

i set it up as x + 1 = A(x^2 + 1) + (Bx + C)x

Answer 3

yep

Answer 4

A-B= 0
A=1
C=1
B=?

Answer 5

so A = 1 but now I'm stuck

Answer 6

if did if x = 0 and got A = 1, but i don't know how to get the other two

Answer 7

coeff. on x^2 terms are A and B.
there are no x^2 terms in your numerator
so A-B must be zero

Answer 8

so rewrite x^2 ( A + B)?

Answer 9

coeff. on your x term is C
there is x in your numerator so C must be 1

Answer 10

sure, if you like..

Answer 11

set it up like this
coeff. on x^n original expansion
x^2 0 A+B
x^1 1 C
x^0 1 A

Answer 12

is this a method you learned? I don't understand why you set it up like this

Answer 13

I mean, I did that for the sake of clarity. To actually solve (most the time), you write equations:
A+B = 0
A= -B
C=1
A=1
B= -1

Answer 14

OH

Answer 15

i see what you did but how do you know A + B = 0

Answer 16

there's no x^2 in your original numerator

Answer 17

ie there's 0*x^2 in your original, in your expansion there's (A+B)*x^2

so A+B must equal 0

Answer 18

thanks

Answer 19

sure.

Answer 20

ok i have 1 more problem

Answer 21

if you don't mind

Answer 22

sure

Answer 23

post a new question

Answer 24

\[\int\limits_{}^{} \frac{ x^2 + x + 3 }{ x^4 + 6x^2 + 9 }\]

Answer 25

dx haha