It's a bit long
If the equation above holds true for positive integers and with least values and being square-free, find .
The answer is 63.
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1 solution
\begin{aligned} I & = \int_0^1 \frac{3x+2}{\sqrt{(x^2+4x+1)^5}} \ dx \\ & = \int_0^1 \frac{3x+2}{(x^2+4x+1)^\frac{5}{2}} \ dx \\ & = \int_0^1 \left( \frac{3}{2} \cdot \frac{2x+4}{(x^2+4x+1)^\frac{5}{2}} - \frac{4}{(x^2+4x+1)^\frac{5}{2}} \right) dx \\ & = - \frac{1}{(x^2+4x+1)^\frac{3}{2}} \bigg|_0^1 - \int_0^1 \frac{4}{((x+2)^2-3)^\frac{5}{2}} \ dx \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9\sqrt{3}} \int_0^1 \frac{1}{\left(\left(\color{#3D99F6}{\frac{x+2}{\sqrt{3}}}\right)^2-1\right)^\frac{5}{2}} \ dx \quad \quad \small \color{#3D99F6}{\text{Let }\sec u = \frac{x+2}{\sqrt{3}} \implies \sec u \tan u \ du = \frac{dx}{\sqrt{3}}} \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \int_{\pi/6}^{\tan^{-1}\sqrt{2}} \frac{\sec u}{\tan^4 u} \ du \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \int_{\pi/6}^{\tan^{-1}\sqrt{2}} \frac{\cos^3 u}{\sin^4 u} \ du \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \int_{1/2}^\sqrt{2/3} \frac{\cos^2 u}{\sin^4 u} \ d\sin u \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \int_{1/2}^\sqrt{2/3} \left( \frac{1}{\sin^4 u} - \frac{1}{\sin^2 u} \right) \ d\sin u \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \left[ -\frac{1}{3\sin^3 u} + \frac{1}{\sin u} \right]_{1/2}^\sqrt{2/3} \\ & = 1 - \frac{1}{6\sqrt{6}} - \frac{4}{9} \left[ -\frac{1}{2}\sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} + \frac{8}{3} - 2 \right] \\ & = \frac{19}{27} - \frac{5}{6\sqrt{6}} \end{aligned}
⟹ A + B + C + D + E = 1 9 + 2 7 + 5 + 6 + 6 = 6 3