∫3cot⁡3π4−xcos2⁡xtan3⁡x+tan2⁡x+tan⁡xdx=6tan−1⁡(f(x))+c then f(x)=

I=∫3cot⁡3π4−xcos2⁡xtan3⁡x+tan2⁡x+tan⁡xdxI=∫3tan⁡x−π4sec2⁡xtan3⁡x+tan2⁡x+tan⁡xdx=∫3(tan⁡x−1)sec2⁡x(1+tan⁡x)tan3⁡x+tan2⁡x+tan⁡xdx=∫31−1tan2⁡xsec2⁡xtan⁡x+2+1tan⁡xtan⁡x+1+1tan⁡xdxlet ,tan⁡x+1+1tan⁡x=t21−1tan2⁡xsec2⁡xdx=2tdt=∫3⋅2t1+t2tdt=6tan−1⁡tan⁡x+1+1tan⁡x+C

∫3cot⁡3π4−xcos2⁡xtan3⁡x+tan2⁡x+tan⁡xdx=6tan−1⁡(f(x))+c then f(x)=

I=∫3cot⁡3π4−xcos2⁡xtan3⁡x+tan2⁡x+tan⁡xdxI=∫3tan⁡x−π4sec2⁡xtan3⁡x+tan2⁡x+tan⁡xdx=∫3(tan⁡x−1)sec2⁡x(1+tan⁡x)tan3⁡x+tan2⁡x+tan⁡xdx=∫31−1tan2⁡xsec2⁡xtan⁡x+2+1tan⁡xtan⁡x+1+1tan⁡xdxlet ,tan⁡x+1+1tan⁡x=t21−1tan2⁡xsec2⁡xdx=2tdt=∫3⋅2t1+t2tdt=6tan−1⁡tan⁡x+1+1tan⁡x+C

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$\int \frac{3 \cot (\frac{3 π}{4} - x)}{\cos^{2} x \sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x = 6 \tan^{- 1} (f (x)) + c then f (x) =$

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$\tan x + \cot x$

$\tan x - \cot x$

$\tan x + \cot x - 1$

$\tan x + \cot x + 1$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{3 \cot (\frac{3 π}{4} - x)}{\cos^{2} x \sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x \\ I = \int \frac{3 \tan (x - \frac{π}{4}) \sec^{2} x}{\sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x \\ = \int \frac{3 (\tan x - 1) \sec^{2} x}{(1 + \tan x) \sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x \\ = \int \frac{3 (1 - \frac{1}{\tan^{2} x}) \sec^{2} x}{(\tan x + 2 + \frac{1}{\tan x}) \sqrt{\tan x + 1 + \frac{1}{\tan x}}} d x \\ let , \tan x + 1 + \frac{1}{\tan x} = t^{2} \end{array}$

$\begin{array}{l} (1 - \frac{1}{\tan^{2} x}) \sec^{2} x d x = 2 t d t \\ = \int \frac{3 \cdot 2 t}{(1 + t^{2}) t} d t \\ = 6 \tan^{- 1} (\tan x + 1 + \frac{1}{\tan x}) + C \end{array}$