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

Explanation

Let `I=int(x-3) sqrt{x^{2}+3 x-18} {dx}`

`x-3=A(frac{d}{d x}(x^{2}+3 x-18))+B`

`x-3=A(2 x+3)+B`

Equating coefficients

`1=2 A` and `-3=3 A+B` or `A=frac{1}{2}` and `B=-frac{9}{2}`

`therefore I=int(frac{1}{2}(2 x+3)-frac{9}{2}) sqrt{x^{2}+3 x-18} {dx}`

`=frac{1}{2} int(2 x+3) sqrt{x^{2}+3 x-18} {dx}-frac{9}{2} int sqrt{x^{2}+3 x-18} {dx}`

`=frac{1}{2} I_{1}-frac{9}{2} I_{2} ldots(1)` Now, `I_{1}=int(2 x+3) sqrt{x^{2}+3 x-18} d x` Let `x^{2}+3 x-18=u`

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