If f(x)=∫ e^x(tan x+tan^2x) dx and f(0)=0 , then f≤ft((π)/4) equals:

The correct answer is Option (3) → 1

Since \( \tan^{2}x=\sec^{2}x-1 \), the integrand splits into a standard form and a simple part.

    \( \displaystyle e^{x}(\tan x+\tan^{2}x)=e^{x}(\tan x+\sec^{2}x)-e^{x} \)

Using \( \int e^{x}\left[g(x)+g'(x)\right]dx=e^{x}g(x) \) with \( g(x)=\tan x \),

    \( \displaystyle f(x)=e^{x}\tan x-e^{x}+C \)

From f(0) = 0,

    \( \displaystyle 0=0-1+C \)

    \( \displaystyle \Rightarrow\ C=1 \)

Then

    \( \displaystyle f\left(\frac{\pi}{4}\right)=e^{\pi/4}-e^{\pi/4}+1 \)

    \( \displaystyle =1 \)