Question

Evaluate: $\int (x-3)\sqrt{x^2+3x-18}dx$

Answer 1

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

Put $x^2+3x-18=t\Rightarrow (2x+3)dx=dt$

Let $x-3=A(2x+3)+B$

Comparing coefficients:

We get $A=\frac{1}{2},B=\frac{-9}{2}$

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

$=\frac{1}{2}\int \sqrt{t}dt-\frac{9}{2}\int \sqrt{x^2+3x+{\left(\frac{3}{2}\right)}^2-18-\frac{9}{4}}dx$

$=\frac{1}{2}\left[\frac{2t^{\frac{3}{2}}}{3}\right]-\frac{9}{2}\int \sqrt{{(x+\frac{3}{2})}^2-{\left(\frac{9}{4}\right)}^2}dx$

$=\frac{1}{3}(x^2+3x-18{)}^{\frac{3}{2}}-\frac{9}{8}(2x+3)\sqrt{x^2+3x-18}+\frac{729}{16}\log |(x+\frac{3}{2})+\sqrt{x^2+3x-18}|+C$