Originally Posted by

**mr fantastic** $\displaystyle \displaystyle I = \int_{x=1}^{x=e} \sqrt{1 + \frac{1}{x^2}} \, dx$.

Substitute $\displaystyle x = e^u$.

Then:

1. $\displaystyle \displaystyle \frac{dx}{du} = e^u \Rightarrow dx = e^u du$.

2. $\displaystyle x = 1 \Rightarrow u = 0$ and $\displaystyle x = e \Rightarrow u = 1$.

3. $\displaystyle \displaystyle \sqrt{1 + \frac{1}{x^2}} = \sqrt{1 + e^{-2u}}$.

Then $\displaystyle \displaystyle I = \int_{u = 0}^{u = 1} \sqrt{1 + e^{-2u}} \, e^u \, du = \int_{u = 0}^{u = 1} \sqrt{e^{2u} + 1} \, du $.

But since it's a definite integral, u is just a dummy variable and you can replace the symbol u with the symbol x.

You need to thoroughly review and revise how to integrate because your instructor obviously assumes you know how to do things like this (otherwise you would not be asked to do questions like this). And most likely you need to review basic algebra (such as index laws etc.) too.