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Evaluate :`int(2x-3)/(x^2+ 3x-18) dx`

Answer» Let `2x-3=lambdad/(dx)(x^(2)+3x-18)+mu`
Then `2x-3=lambdad/(dx)(x^(2)+3x-18)+mu`
Comparing the coefficients of like power of x, we get.
`2lambda=2`, and `3lambda+mu=-3 rArr lambda=1` and `mu=-6`
So, `int(2x-3)/(x^(2)+3x-18)dx = int(2x+3-6)/(x^(2)+3x+9/4-9/4-18dx)= int(2x+3)/(x^(2)+3x-18)dx - 6int1/(x^(2)+3x-18)dx`
`=ln|x^(2)+3x-18|-6int1/(x^(2)+3x+9/4-9/4-18)dx = ln|x^(2)+3x-18|-6int1/((x+3/2)^(2)-(9/2)^(2))dx`
`ln|x^(2)+3x-18|-6.1/(2(9/2)ln|(x+3/2-9/2)/(x+3/2+9/2)|+C) = ln|x^(2)+3x-18|-2/3ln|(x-3)/(x+6)|+C`


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