Improper Integral Solution Check: Is Your Answer Accurate?

B18
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Hi guys just want to check my answer for the following improper integral.

∫(2 to ∞) dv/v^2+2v-3.

After doing partial fractions, integrating and evaluating I got the following for the answer: 0-(1/4)ln(1/5)

How does this compare to other answers?

Is there a way I can accurately check this answer myself?
Thanks!
 
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if I'm not mistaken u= (v+1) is pretty easy and it ends up fitting the arctanh rule. Seems that it'll involve an inverse hyperbolic function
 
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What exactly do you mean that substituting u=v+1 is easy? I don't see any substitution in this problem
 
B18 said:
What exactly do you mean that substituting u=v+1 is easy? I don't see any substitution in this problem

I meant completing the square so that the denominator is (v+1)^2 -4. Multiply by -1/-1 and you have -dv/(4-(v+1)^2)
which fits the rule ∫du/(a^2-u^2)=1/2a(ln (a+u/a-u)+c
 
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Alright, that would make sense. Hopefully people get a similar final answer.
 
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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