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 = \(\int\frac{dt}{1-t^2}\) (\(\because t = tan x\), ⇒ sec2x dx = dt)

 = \(\frac12log|\frac{1+t}{1-t}|+c\)

 = \(\frac12log|\frac{1+tan x}{1-tanx}|+c\) 

 = \(\frac12log|tan(\frac {\pi}4+x)|+c\)

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