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

Question

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

Infinity Learn · Accepted Answer

First show that $$f(x)=2x$$. Now , $$I=$$∫$$f(x)$$$$tan$$−$$1$$⁡$$x1+x22dx=$$∫$$2x1+x22tan$$−$$1$$⁡xdx Put $$x=$$$$tan$$⁡θ , so $$thatI=$$∫$$2tan$$⁡θ$$sec4$$⁡θθ$$sec2$$⁡θdθ$$=$$∫θ$$sin$$⁡$$(2$$θ$$)d$$θ$$=$$−$$12$$θ$$cos$$⁡$$2$$θ$$+14sin$$⁡$$2$$θ$$+C=C$$−$$12$$θ$$1$$−$$tan2$$⁡θ$$1+tan2$$⁡θ$$+12tan$$⁡θ$$1+tan2$$⁡θ$$=C$$−$$121$$−$$x21+x2tan$$−$$1$$⁡$$x+12x1+x2=C$$−$$11+x2tan$$−$$1$$⁡$$x+12tan$$−$$1$$⁡$$x+12x1+x2$$

Let $f$ be a continuous function satisfying
$f (x + y) = f (x) + f (y),$ for each $x, y \in R$
and $f (1) = 2$ then $\int \frac{f (x) \tan^{- 1} x}{{(1 + x^{2})}^{2}} d x$ is equal to

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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

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Let $f$ be a continuous function satisfying
$f (x + y) = f (x) + f (y),$ for each $x, y \in R$
and $f (1) = 2$ then $\int \frac{f (x) \tan^{- 1} x}{{(1 + x^{2})}^{2}} d x$ is equal to