TITLE: Definite Integration having d/dx operator inside the integral sign .
QUESTION [0 upvotes]: $$\int_{-1}^1 \frac{d}{dx}\ \frac{1}{1 + 2^{1/x}}\ dx$$
How to approach this integral ?
Should I differentiate the expression and then integrate ? 
I tried that but couldn't proceed 
Thank you.

REPLY [1 votes]: By using the integration by parts we can solve it in a very easy way as Dog_69 pointed out.
$$\int_1^{+1} \frac{d}{dx} \frac{1}{1 + 2^{1/x}}\ dx = \frac{1}{1 + 2^{1/x}}\bigg|_{-1}^0 + \frac{1}{1 + 2^{1/x}}\bigg|_0^{1}$$
The result is simply
$$\frac{1}{1 + 2} = \frac{2}{3}$$
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