Answer
See step by step
Work Step by Step
We know that
$$\int \frac{1}{1+x^2}dx = tan^{-1}(x)$$
Therefore,
$$\int_0^{1/x} \frac{1}{1+t^2}dt =tan^{-1}(t)]_0^{1/x} = tan^{-1}(1/x)$$
And,
$$\int_0^{x} \frac{1}{1+t^2}dt =tan^{-1}(t)]_0^{x} = tan^{-1}(x)$$
So,$$ f(x) = \int_0^{1/x} \frac{1}{1+t^2}dt + \int_0^{x} \frac{1}{1+t^2}dt = tan^{-1}(x) + tan^{-1}(1/x)$$
Let x = tan(y).
Therefore, $\frac{1}{x} = cot(y) = tan(\pi/2 - y)$
Thus, $$tan^{-1}(x) = y$$
and. $$tan^{-1}(1/x) =\pi/2 - y$$
So, $$f(x) = tan^{-1}(1/x)+tan^{-1}(x) = y+(\pi/2-y)$$
Or, $$f(x) = \pi/2$$
Hence, f(x) is constant and equal to $\pi/2$ for x>0
