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sec-2-x-tan-x-1-4-tan-x-2-dx-




Question Number 86269 by M±th+et£s last updated on 27/Mar/20
∫((sec^2 (x))/((tan(x)−1)^4 (tan(x)−2))) dx
$$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)−\mathrm{1}\right)^{\mathrm{4}} \left({tan}\left({x}\right)−\mathrm{2}\right)}\:{dx} \\ $$
Commented by john santu last updated on 28/Mar/20
u = tan x−2   ⇒ ∫ (du/((u+1)^4 (u)))   decomposition  (1/((u+1)^4 (u))) = (p/(u+1))+(q/((u+1)^2 ))+(r/((u+1)^3 ))  + (s/((u+1)^4 )) + (t/u)
$${u}\:=\:\mathrm{tan}\:{x}−\mathrm{2}\: \\ $$$$\Rightarrow\:\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}} \left({u}\right)}\: \\ $$$${decomposition} \\ $$$$\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}} \left({u}\right)}\:=\:\frac{{p}}{{u}+\mathrm{1}}+\frac{{q}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{r}}{\left({u}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$+\:\frac{{s}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}} }\:+\:\frac{{t}}{{u}} \\ $$$$ \\ $$

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