Integrating Using Partial Fractions

Jay9313
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Homework Statement


This is an arc length problem in three dimensions. I was given the vector r(t)=<et, 1, t> from t=0 to t=1


Homework Equations


Arc Length= \int |\sqrt{r&#039;(t)}| dt from t1 to t2
where |\sqrt{r&#039;(t)}| is the magnitude of the derivative of the vector

The Attempt at a Solution



I took the derivative and got the magnitude and simplified it down to
∫ √(e2t+1) dt
I then set u=e2t+1
I then simplified and substituted until I got to:∫\frac{\sqrt{(u)}}{u-1} du

My professor said to use partial fractions from here, but I'm not sure how to do that.
 
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Try one more substitution to get rid of the square root on top first. Then do partial fractions.
 
That's not really working at all
 
What did you try?
 
I split it up, and it made it worse, and I had to do long division, but I got an answer. It's nasty, but I got an answer.
 
Presumably, you ended up with something like ##\frac{v^2}{v^2-1}## after the substitution. A good technique to avoid doing long division is to add and subtract judiciously:
$$\frac{v^2}{v^2-1} = \frac{(v^2-1)+1}{v^2-1} = 1 + \frac{1}{v^2-1}.$$
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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