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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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