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

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

= 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?

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

Note first post second line should be x+5)/(x^3−9x) = A/x + B/(x-3) + C/(x+3)

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

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

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