$Q.$ Evaluate the following integral :
$\int_{1}^{2}\frac{x+\tan x}{x+\sin x}dx$. Numerically I found that the answer is roughly $1.000006$ but I am unable to compute using the analytic methods.
I tried first computing by splitting:
$\int_{1}^{2}\frac{x}{x+\sin x}dx+\int_{1}^{2}\frac{\tan x}{x+\sin x}dx$
and then applying by-parts to each of them, but that results in a very difficult task.
Hint
First of all, I do not think that the antiderivative can be found analytically.
Second, there is a problem at $x=\frac{\pi}{2}$ because of the tangent.
So, as suggested by André Nicolas, consider $$\int_{1}^{2}\frac{x+\tan x}{x+\sin x}dx=\int_{1}^{\frac{\pi}{2}-\epsilon}\frac{x+\tan x}{x+\sin x}dx+\int_{\frac{\pi}{2}+\epsilon}^{2}\frac{x+\tan x}{x+\sin x}dx$$ and look at the limits when $\epsilon$ goes to $0$.
