If ∫$$tan$$⁡$$x1+$$$$tan$$⁡$$x+tan2$$⁡$$xdx=x$$−KAtan−$$1$$⁡Ktan⁡$$x+1A+C$$,(C$$ is a constant of $$integration), then the ordered pair (K$$,$$A) is equal to

Question

If ∫$$tan$$⁡$$x1+$$$$tan$$⁡$$x+tan2$$⁡$$xdx=x$$−KAtan−$$1$$⁡Ktan⁡$$x+1A+C$$,(C$$ is a constant of $$integration),  then the ordered pair (K$$,$$A) is equal to

Infinity Learn · Accepted Answer

Let $$I=$$∫$$tan$$⁡$$x1+$$$$tan$$⁡$$x+tan2$$⁡xdx⇒$$I=$$∫$$tan$$⁡$$x+1+tan2$$⁡xtan⁡$$x+1+tan2$$⁡xdx−∫$$1+tan2$$⁡$$x1+$$$$tan$$⁡$$x+tan2$$⁡xdx⇒$$I=x$$−∫$$sec2$$⁡$$x1+$$$$tan$$⁡$$x+tan2$$⁡xdx Put $$tan$$⁡$$x=t$$⇒$$sec2$$⁡x⋅$$dx=dt$$∴$$I=x$$−∫$$dtt2+t+1$$∴$$I=x$$−∫$$dtt2+t+14+1$$−$$14=x$$−∫$$dtt+122+322$$⇒$$I=x$$−$$23tan$$−$$1$$⁡$$t+1232+C$$⇒$$I=x$$−$$23tan$$−$$1$$⁡$$2tan$$⁡$$x+13+C$$∴$$A=3$$ and $$K=2$$

$If \int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x = x - \frac{K}{\sqrt{A}} \tan^{- 1} (\frac{K \tan x + 1}{\sqrt{A}}) + C, (C is a constant of integration),$ $then the ordered pair (K, A) is equal to$

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If ∫tan⁡x1+tan⁡x+tan2⁡xdx=x−KAtan−1⁡Ktan⁡x+1A+C,(C is a constant of integration), then the ordered pair (K,A) is equal to

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$If \int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x = x - \frac{K}{\sqrt{A}} \tan^{- 1} (\frac{K \tan x + 1}{\sqrt{A}}) + C, (C is a constant of integration),$ $then the ordered pair (K, A) is equal to$