Evaluate - - xtan x- 2x-1dx-

The correct option is B 1/2t√(t^2+1)+1/2log [ t+√(t^2+1) ]. where t= x Let #160;I=∫ #160; #160;xtan #160;x√( ^ 2 x+1dx ) Put secx=t ...

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Integration by Substitution

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Question

Evaluate : $\int sec x tan x \sqrt{{sec}^{2} x + 1} d x .$

\frac{1}{2} t \sqrt{t^{2} + 1} - \frac{1}{2} log [t + \sqrt{t^{2} + 1}] .

where

t = sec x

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\frac{1}{2} t \sqrt{t^{2} + 1} + \frac{1}{2} log [t + \sqrt{t^{2} + 1}] .

where

t = sec x

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t \sqrt{t^{2} + 1} + log [t + \sqrt{t^{2} + 1}] .

where

t = sec x

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\frac{1}{2} t \sqrt{t^{2} + 1} + \frac{1}{2} log [t - \sqrt{t^{2} + 1}] .

where

t = sec x

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