If ∫1/((sin x+4)(sin x - 1))dx=A1/(tanx/2-1)+Btan^-1(f(x))+C, then:

Given,

$\int\frac{1}{(\sin x+4)(\sin x - 1)}dx=A\frac{1}{\tan\frac{x}{2}-1}+B\tan^{-1}(f(x))+C$

$\int\frac{1}{(\sin x+4)(\sin x - 1)}dx=\frac{1}{5}\int\frac{(\sin x+4)-(\sin x - 1)}{(\sin x+4)(\sin x - 1)}dx=\frac{1}{5}\int\frac{1}{\sin x - 1}dx-\frac{1}{5}\frac{1}{\sin x +4}dx$ $=\frac{1}{5}\int\frac{2\,dt}{2t-1-t^2}-\frac{1}{5}\int\frac{2\,dt}{2t+4(1+t^2)}$ [Putting $\tan\frac{x}{2}=t$]

$=-\frac{2}{5}\int\frac{dt}{t^2-2t+1}-\frac{1}{10}\int\frac{dt}{t^2+\frac{1}{2}t+1}=-\frac{2}{5}\int\frac{1}{(t-1)^2}dt-\frac{1}{10}\int\frac{1}{(t+\frac{1}{4})^2+(\frac{\sqrt{15}}{4})^2}dt$

$=\frac{2}{5}\frac{1}{(t-1)}-\frac{2}{5\sqrt{15}}\tan^{-1}(\frac{4t+1}{\sqrt{15}})+C=\frac{2}{5}.\frac{1}{\tan\frac{x}{2}-1}-\frac{2}{5\sqrt{15}}\tan^{-1}(\frac{4\tan\frac{x}{2}+1}{\sqrt{15}})+C$

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