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## Homework Statement

Find the length of the curve:

##x=\frac{t}{1+t}##

##y=\ln(1+t)##

where ##0 \leq t \leq 2##

Length of curve integral ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt##

## Homework Equations

##(\frac{dx}{dt})^2 = \frac{1}{(1+t)^4}## ##(\frac{dy}{dt})^2 = \frac{1}{(1+t)^2} ##

## The Attempt at a Solution

So, I think somewhere I missed a more efficient technique to use (or made an algebra mistake which is most likely), because otherwise I think this integral gets crazy too fast for a simple problem.

Start with ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt## to which when we substitute our given problem, we get ##\int^2_0 \sqrt{\frac{1}{(1+t)^4}+\frac{1}{(1+t)^2}} dt## = ##\int^2_0 \sqrt{\frac{(1+t)^2+1}{(1+t)^4}} dt ## = ##\int^2_0 \frac{1}{(1+t)^2} \sqrt{(1+t)^2+1} dt##

To try to make this simpler, I said well, just let ##u = 1+t## therefore ##du=dt## so our integral looks like

##\int^2_0 \frac{1}{(u)^2} \sqrt{(u)^2+1} du## which then I believe calls for a trig substitution (let ##u=tan\theta##)... but that seems crazy! At this point why wouldn't I just de-parametrize the equation from the start if I wanted a crazy integral.

If anyone has a simpler way to go about this integral, I'd love to know. Otherwise if this seems to be the correct path, I'll just keep chugging along!

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