Question

Determine the following intergral by using the partial fractions method. (show complete workings out ) (i dont understand how to get the correct answer) \int\limits_{}^{}(x+5)/(x^3-9x) dx

Answer 1

\[\int\limits\limits_{}^{}(x+5)/(x^3-9x) dx\]

Answer 2

= x^3−9x = x(x-3)(x+3)

(x+5)/(x^3−9x = A/x + B/(x-3) + C(x+3)
Multiply thru by x^3−9x

x+5 = A(x^2-9) + B(x(x+3)) + C(x(x-3))
this is an identity so its true for all values of x and we can find the values of A,B and C by using different values ox or by equating coefficients

nedd any more help from here?

Answer 3

Ok

let x = 0
5 = -9A
A = -5/9
equate coefficients of x:
1 = 3B - 3C
equate cofficients of x^2:
0 = (-5/9) + B + C
solving equatiuions simultaneously
3B + 3C = 15/9
3B - 3C = 1
subtract:
6C = 6/9
C = 1/9
B = 5/9 - 1/9 = 4/9

Answer 4

Note first post second line should be

x+5)/(x^3−9x) = A/x + B/(x-3) + C/(x+3)

Answer 5

so the integral becomes:

INT ( -5/9x + 4/9(x+3) + 1/9(x-3)) dx

= -(5/9) ln x + (4/9) ln (x+3) + (1/9) ln(x-3) + C

= 1/9[4 ln (x+3) - 5 lnx + ln(x-3) + C

Answer 6

this further simplifies to

1/9{ [ln(x+3)^4(x-3)]/lnx^5}