Partial Fractions, Irreducible quadratic factors

rocomath
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Arc Length, Irreducible quadratic factors

i'm having a hard time seeing this method, and i have to use this method on one of the problems I'm doing to find it's Arc Length.

L=\int_{\sqrt{2}}^{\sqrt{1+e^{2}}}\frac{v^{2}dv}{v^{2}-1}}

the book suggests to first divide then use a u-substitution. i know that when the power in the numerator is greater than the denominator, i can perform long-division, but i don't see how i can divide this.
 
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i don't know whether i am gettin u well, but you could do this inegral by doing this to the function under the integral sign

v^2/(v^2-) =(v^2-1+1)/(v^2-1) = (1- 1/(v-1)(v+1) )

then you could use partial fraction by letting

1/(v-1)(v+1) = A/(v-1) + B /(v+1), and then finding the values of B and A. Pardone me if i misunderstood u.
 
Well, like sutupidmath did, add one and then subtract one to the numerator. Then, instead of factoring the new denominator and partial fractions-ing, Try a hyperbolic substitution, I think you will find it quicker.
 
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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