$\displaystyle L = \int_{0}^{\frac{\pi}{4}} 2\sqrt{1+t^2} dt \rightarrow 2\int_{0}^{\frac{\pi}{4}} \sqrt{1^2 +t^2} dt$

$\displaystyle t = tan(\theta) \rightarrow dt = sec^2(\theta) d\theta$

$\displaystyle = 2 \int \sqrt{1 + tan^2(\theta)} * sec^2(\theta) d\theta \rightarrow 2\int sec^3(\theta) d\theta$

$\displaystyle = 2 \int (1+tan^2(\theta)) * sec(\theta) d\theta \rightarrow 2\int sec(\theta) + sec(\theta)tan^2(\theta) d\theta$

$\displaystyle =2\int sec(\theta) d\theta + 2\int sec(\theta)tan^2(\theta) d\theta$

$\displaystyle =2 \int sec(\theta) d\theta \rightarrow 2\ln[sec(\theta) + tan(\theta)]$

To simplify the first integral answer isn't it just setting up a right triangle and figuring out what the values will be? I got $\displaystyle sec(\theta) = \sqrt{t^2 +1}$ and $\displaystyle tan(\theta) = t$

So then the answer to that first integral would be $\displaystyle 2\ln[\sqrt{t^2+1} + t]$ right? To be honest I'm really lost on this entire problem after I got half way through the integral. How do I finish this problem up exactly? And when exactly do I change the limits on this? Assuming that I will have to at some point. I attached what wolfram got as an answer as well for the indefinite integral for a reference. I'm not reallly familiar with the reduction formula as I never learned that in calculus 2. Perhaps I also went too far in the integral above after looking at wolfram's work, though I'm not too sure again. Any help appreciated.

*Only put "Supposedly" in title because I wasn't completely sure if integration by parts is in this problem or not. But after submitting it seems more likely that it is.