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Question Number 179066 by cortano1 last updated on 24/Oct/22
     If f(x)=∫ (x^2 /(x^2 +tan x)) dx       then ∫ ((tan x)/(x^2 +tan x)) dx =?
$$\:\:\:\:\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\:\:\:\:\mathrm{then}\:\int\:\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$
Answered by Ar Brandon last updated on 24/Oct/22
f(x)=∫(x^2 /(x^2 +tanx))dx=∫(((x^2 +tanx)−tanx)/(x^2 +tanx))dx           =∫(1−((tanx)/(x^2 +tanx)))dx=x−∫((tanx)/(x^2 +tanx))dx          ⇒∫((tanx)/(x^2 +tanx))dx=x−f(x)
$${f}\left({x}\right)=\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{tan}{x}}{dx}=\int\frac{\left({x}^{\mathrm{2}} +\mathrm{tan}{x}\right)−\mathrm{tan}{x}}{{x}^{\mathrm{2}} +\mathrm{tan}{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\int\left(\mathrm{1}−\frac{\mathrm{tan}{x}}{{x}^{\mathrm{2}} +\mathrm{tan}{x}}\right){dx}={x}−\int\frac{\mathrm{tan}{x}}{{x}^{\mathrm{2}} +\mathrm{tan}{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\int\frac{\mathrm{tan}{x}}{{x}^{\mathrm{2}} +\mathrm{tan}{x}}{dx}={x}−{f}\left({x}\right) \\ $$

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