Quick question on tables and integration

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trajan22
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ok so here is the problem

\int\frac{dx}{\sqrt(x^2-4x)}

the table integral I am supposed to use is this
\int\frac{du}{\sqrt{u^2-a^2}}=ln(u+\sqrt{u^2-a^2}+C

Is it proper to make my u=x^2 and a=2x^(1/2)
I am asking because the solution guide tells me to complete the square, and then proceed to pick the u and a
 
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No. a is assumed to be constant.
 
so if a is any real number i can do this, but if any nonconstant variable is there than I need to find another way? Sorry it has been a while since I've done this.
 
It doesn't matter whether it's "a" or "b", "y", "z",...You integrate wrt "x" and that's all that matters. Don't use tables of integrals, when the integrations can be done explicitely.
 
dextercioby said:
It doesn't matter whether it's "a" or "b", "y", "z",...You integrate wrt "x" and that's all that matters. Don't use tables of integrals, when the integrations can be done explicitely.

Yes, you're right. I was being lazy...constant in this context means a does not depend on x.

Another one of the little nuances in math.
 
I wouldn't consider the definition of "constant" to be a "little nuance"!
 
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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