$$\int_{-1}^1 \frac{d}{dx}\ \frac{1}{1 + 2^{1/x}}\ dx$$
How to approach this integral ?
Should I differentiate the expression and then integrate ?
I tried that but couldn't proceed
Thank you.
$\endgroup$ 02 Answers
$\begingroup$Since the fundament calculus is not applicable here, derive the expression inside the integral first, which is trivial and gives you:
$$\frac{d}{dx} \frac{1}{1 + 2^{1/x}} = \frac{2^{1/x} \ln (2)}{\left(2^{1/x}+1\right)^2 x^2}$$
Then integrate this from $-1$ to $1$, which is another simple integration which gives you
$$\frac{2}{3}$$
$\endgroup$ 7 $\begingroup$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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