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Calculus Level pending

$\large \int \frac {dx}{\sqrt {2-3x-x^2}} = \arcsin \left(\frac {Ax+B}{\sqrt D} \right) +C$

The equation above holds true for positive integers $A$ , $B$ and $D$ with $D$ being square-free, and $C$ as the arbitrary constant of integration . Find $A+B+D$ .

25 20 21 22 5

1 solution

Chew-Seong Cheong
Mar 1, 2017

$\begin{aligned} I & = \int \frac {dx}{\sqrt{2-3x-x^2}} \\ & = \int \frac {dx}{\sqrt{\frac {17}4 - \left(x+\frac 32 \right)^2}} \\ & = \int \frac {dx}{\frac {\sqrt{17}}2 \sqrt{1 - \left({\color{#3D99F6}\frac {2x+3}{\sqrt{17}}} \right)^2}} & \small \color{#3D99F6} \text{Let } \sin \theta = \frac {2x+3}{\sqrt{17}}, \ \cos \theta \ d \theta = \frac 2{\sqrt{17}} dx \\ & = \int \frac {\cos \theta \ d\theta}{\sqrt{1 - {\color{#3D99F6}\sin^2 \theta} }} \\ & = \int \frac {\cos \theta \ d\theta}{\cos \theta} \\ & = \int 1 \ d\theta \\ & = \theta + C \\ & = \arcsin \left(\frac {2x+3}{\sqrt{17}}\right) + C \end{aligned}$

$\implies A+B+D = 2+3+17 = \boxed{22}$

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