How do I solve this integral!?

Question

I'm trying to solve this partial fraction integral, int:10/((x-1)(x^2 +9)) dx
I wrote that A/(x-1) + (Bx+C)/(x^2+9)= 10/((x-1)(x^2 +9))
I multiplied across the equation by (x-1)(x^2+9) and got
A(x^2 +9) + (Bx+C)(x-1) = 10 Then I multiplied the left side of the equation out and got..
Ax^2 +9A + Bx^2 -Bx + Cx -C = 10 Then I set up a system of linear equations like this
A +B = 0
-B +C = 0
9A -C = 10
Now I'm stuck.. I may have made a mistake somewhere, but in any case I'm not sure how to solve this system of linear equations. What should I do?

Answer 1

you can solve the three equations as volterwd and eric have shown.

but an easier way will be to put x =1 in
A(x^2 +9) + (Bx+C)(x-1) = 10 to get 10A = 10 or A =1

then equate coefficients of x^2 to get A+B = 0
then use A =1 to get B = -1

compare constant terms to get 9A - C = 10
then use A = 1 to get C = -1

Answer 2

To solve the system, first add the first two equations:

A + B = 0
-B + C = 0

A + C = 0

Then add this result to the third equation:

A + C = 0
9A - C = 10
10A = 10

A = 1
B = -1
C = -1

so, 10/((x-1)(x^2 +9)) = 1/(x - 1) - (x + 1)/(x^2 + 9)

Answer 3

Without doing a formal method

(1) => A= -B

(2) with (1) 0=-B+C = A+C

=> A=-C

(3) with (2) 10= 9A-C = 9A+A=10A

=> A=1

=> B=C = -1

Not formal but its simple enough

Answer 4

Two words:
Gaussian Elimination


Doug

Answer 5

If you have three equations and three variables, you can solve.
Plug and chug!