How does the int get from that step to this step?

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Book ex: We were given integral of \frac{sqrt of x + 4}{x}/x dx

with the substitution u = sqrt{x + 4}
--> u^2 = x + 4 and x = u^2 - 4
--> dx = 2u du

and the int becomes 2*int of \frac{u^2}{u^2 - 4}du

But how does it go from that to the next step:

2 int of 1 + \frac{4}{u^2 - 4}du ?
 
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You add and subtract 4 to the numerator and regroup.

(u^2+4-4)/(u^2 -4) = (u^2-4)/(u^2-4) + 4/(u^2-4) = 1 + 4/(u^2-4)
 
i'm really confused as to what your original problem is

\int\frac{\frac{\sqrt{x+4}}{x}}{x}}dx

that?
 
is the integral of sqrt (x+4)/x dx -- sorry for the confusion... And how can the book not explain something like that?!? If i see james stewart on the street I am going to fling pebles at him man.
 
i actually like his textbook, what ch./sect. is this?
 
frasifrasi said:
is the integral of sqrt (x+4)/x dx -- sorry for the confusion... And how can the book not explain something like that?!? If i see james stewart on the street I am going to fling pebles at him man.

:smile: Pebbles?! I guess that'll learn him.
 
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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