Definite Integration having d/dx operator inside the integral sign .

Question

$$\int_{-1}^1 \frac{d}{dx}\ \frac{1}{1 + 2^{1/x}}\ dx$$

How to approach this integral ?

Should I differentiate the expression and then integrate ?

I tried that but couldn't proceed

Thank you.

Answer 1

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

$$\frac{2}{3}$$

Answer 2

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

$$\frac{1}{1 + 2} = \frac{2}{3}$$