help plz !!!

integration of sec x log(sec x + tan x) dx 

Ans : I =  ⌡ secx*log(secx+tanx) dx

Let u=log(secx+tanx)

Differentiate wrt  "x"

du/dx = {1/(secx+tanx)} * { d(secx)/dx + d(tanx)/dx}

du/dx = {1/(secx+tanx)} * { (secx*tanx)+(sec²x)}

du/dx = {1/(secx+tanx)} * { (tanx)+(secx)} * secx

du = secx dx

Therefore: I = ⌡u du

I = (u²/2) + C

I = ( [log(secx+tanx)] ²/2) + C

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