∫ t.dt/1+t^2 ; t ∈ [e^-1, tanx] + ∫ dt/t(1+t^2) ; t ∈ [e^-1, cos tx] =

Question

\(\int\limits_{e^{-1}}^{tanx} \frac{t.dt}{1+ t^2} + \int\limits_{e^{-1}}^{costx} \frac{dt}{t(1+t^2)} =\)

(a) 1

(b) e

(c) e^–1

(d) π/4

Answer 1

Correct option is (a) 1