Indefinite IntegrationHard

Question

If dx = f(x) + A  ln |x + | + C, then

Options

A.f(x) = tan-1 x, A = - 1
B.f(x) = tan-1 x, A = 1
C.f(x) = 2 tan-1 x, A = - 1
D.f(x) = 2 tan-1 x, A = 1

Solution

   Put x = tan θ     ⇒   dx = sec2θ dθ
⇒  I = . sec2θ dθ = θ ∫tanθ secθdθ - ∫1. secθ dθ = θ sec θ -  ln |sec θ + tan θ| + C
=  tan-1x - log |x + | + C

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