Question

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

Answer 1

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

Here,

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

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

On equating,

$2A=1$ and $3A+B=-3$

$A=\frac{1}{2}$ and $B=-\frac{9}{2}$

Thus,

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

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

$I=\frac{1}{2}{I}_1-\frac{9}{2}{I}_2$

$I_1=\frac{1}{2}\int (2x+3)\sqrt{x^2+3x-18}dx$

Put $x^2+3x-18=t$

$(2x+3)dx=dt$

$I_1=\int t^{1/2}dt$

$I_1=\frac{2}{3}{t}^{3/2}+C_1$

$I_1=\frac{2}{3}(x^2+3x-18{)}^{3/2}+C_1$

$I_2=\int \sqrt{x^2+3x-18}dx$

$I_2=\int \sqrt{(x+\frac{3}{2}{)}^2-(\frac{9}{2}{)}^2}dx$

$I_2=\frac{(x+\frac{3}{2})}{2}\sqrt{x^2+3x-18}-\frac{81}{8}\log |(x+\frac{3}{2})+\sqrt{x^2+3x-18}|+C_2$

$I_2=\frac{2x+3}{4}\sqrt{x^2+3x-18}-\frac{81}{8}\log |\frac{2x+3}{2}+\sqrt{x^2+3x-18}|+C_2$

Putting these values, we get,

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

where, $C=\frac{C_1}{2}-\frac{9C_2}{2}$