∫tan(x−α)tan(x+α)tan2xdx=

I=∫tan⁡(x−α)tan⁡(x+α)tan⁡2xdx_ Now tan⁡2x=tan⁡(x+α+x−α)tan⁡2x=tan⁡(x+α)+tan⁡(x−α)1−tan⁡(x+α)tan⁡(x−α)tan⁡2x(1−tan⁡(x+α)tan⁡(x−α)=tan⁡(x+α)+tan⁡(x−α)∴tan⁡(x−α)tan⁡(x+α)tan⁡2x=tan⁡2x−tan⁡(x+α)−tan⁡(x−α)⇒I=∫tan⁡(x−α)tan⁡(x+α)tan⁡2xdx=∫{tan⁡2x−tan⁡(x+α)−tan⁡(x−α)}dx=12log⁡|sec⁡2x|−log⁡|sec⁡(x+α)|−log⁡|sec⁡(x−α)|+C