Let f be a continuous function satisfying f(x+y)=f(x)+f(y), for each x,y∈Rand f(1)=2 then ∫f(x)tan−1⁡x1+x22dx is equal to

First show that f(x)=2x. Now , I=∫f(x)tan−1⁡x1+x22dx=∫2x1+x22tan−1⁡xdx Put x=tan⁡θ , so thatI=∫2tan⁡θsec4⁡θθsec2⁡θdθ=∫θsin⁡(2θ)dθ=−12θcos⁡2θ+14sin⁡2θ+C=C−12θ1−tan2⁡θ1+tan2⁡θ+12tan⁡θ1+tan2⁡θ=C−121−x21+x2tan−1⁡x+12x1+x2=C−11+x2tan−1⁡x+12tan−1⁡x+12x1+x2