\(\int {\frac{{dx}}{{(\sin x + 4)(\sin x - 1)}} = \frac{A}{{\tan \frac{x}{2} - 1}} + B{{\tan }^{ - 1}}} \left( {f(x)} \right) + c\), then

\(A = \frac{2}{5}\), \(B = -\frac{2}{{\sqrt {15} }}\), \(f(x) = \frac{{4\tan \left( {\frac{x}{2}} \right) + 1}}{{\sqrt {5} }}\)

\(A = \frac{1}{5}\), \(B = -\frac{2}{{5\sqrt {15} }}\), \(f(x) = \frac{{4\tan x + 3}}{{\sqrt {15} }}\)

\(A = \frac{-1}{5}\), \(B = \frac{1}{{\sqrt {15} }}\), \(f(x) = \frac{{4\tan \left( {\frac{x}{2}} \right) + 1}}{{\sqrt {15} }}\)

\(A = \frac{2}{5}\), \(B = -\frac{2}{{5\sqrt {5} }}\), \(f(x) = \frac{{4\tan x + 1}}{{\sqrt {5} }}\)

\(\int {\frac{{dx}}{{(\sin x + 4)(\sin x - 1)}} = \frac{A}{{\tan \frac

1.	\(\int {\frac{{dx}}{{(\sin x + 4)(\sin x - 1)}} = \frac{A}{{\tan \frac{x}{2} - 1}} + B{{\tan }^{ - 1}}} \left( {f(x)} \right) + c\), then
A.	\(A = \frac{1}{5}\), \(B = -\frac{2}{{5\sqrt {15} }}\), \(f(x) = \frac{{4\tan x + 3}}{{\sqrt {15} }}\)
B.	\(A = \frac{-1}{5}\), \(B = \frac{1}{{\sqrt {15} }}\), \(f(x) = \frac{{4\tan \left( {\frac{x}{2}} \right) + 1}}{{\sqrt {15} }}\)
C.	\(A = \frac{2}{5}\), \(B = -\frac{2}{{5\sqrt {5} }}\), \(f(x) = \frac{{4\tan x + 1}}{{\sqrt {5} }}\)
D.	\(A = \frac{2}{5}\), \(B = -\frac{2}{{\sqrt {15} }}\), \(f(x) = \frac{{4\tan \left( {\frac{x}{2}} \right) + 1}}{{\sqrt {5} }}\)
Answer» D. \(A = \frac{2}{5}\), \(B = -\frac{2}{{\sqrt {15} }}\), \(f(x) = \frac{{4\tan \left( {\frac{x}{2}} \right) + 1}}{{\sqrt {5} }}\)

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\(\int {\frac{{dx}}{{(\sin x + 4)(\sin x - 1)}} = \frac{A}{{\tan \frac{x}{2} - 1}} + B{{\tan }^{ - 1}}} \left( {f(x)} \right) + c\), then

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