Mathematics

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


SOLUTION
$$I=\int (x-3) \sqrt{x^2+3x-18}dx$$
Here,
$$x-3=A \dfrac{d}{dx}(x^2+3x-18)+B$$

$$x-3=A(2x+3)+B$$
On equating,
$$2A=1$$ and $$3A+B=-3$$
$$A=\dfrac{1}{2}$$ and $$B=-\dfrac{9}{2}$$

Thus,
$$I=\int [\dfrac{1}{2}(2x+3)-\dfrac{9}{2}]\sqrt{x^2+3x-18}dx$$

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

$$I=\dfrac{1}{2}I_{1}-\dfrac{9}{2}I_{2}$$

$$I_{1}=\dfrac{1}{2} \int (2x+3)\sqrt{x^2+3x-18} \ dx$$
Put $$x^2+3x-18=t$$
$$(2x+3) \ dx =dt$$
$$I_{1} = \int t^{1/2} \ dt$$

$$I_{1} = \dfrac{2}{3} t^{3/2}+C_{1}$$

$$I_{1}=\dfrac{2}{3}(x^2+3x-18)^{3/2}+C_{1}$$

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

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

$$I_{2}= \dfrac{(x+\dfrac{3}{2})}{2} \sqrt{x^2+3x-18} -\dfrac{81}{8} \log |(x+\dfrac{3}{2})+\sqrt{x^2+3x-18}|+C_{2}$$

$$I_{2}=\dfrac{2x+3}{4}\sqrt{x^2+3x-18}-\dfrac{81}{8} \log |\dfrac{2x+3}{2}+\sqrt{x^2+3x-18}|+C_{2}$$

Putting these values, we get,
$$I=\dfrac{1}{3}(x^2+3x-18)^{3/2}-\dfrac{9}{8}(2x+3)\sqrt{x^2+3x-18}+\dfrac{729}{16}\log |\dfrac{2x+3}{2}+\sqrt{x^2+3x-18}|+C$$

where, $$C=\dfrac{C_{1}}{2}-\dfrac{9C_{2}}{2}$$
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