Help with integrating arctan function

ssmith147
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Homework Statement


INTEGRATE dx / (2 * root(x)) * (1 + x)



Homework Equations


That's pretty much it!


The Attempt at a Solution


I received this question on a Calc II exam, so I'm only looking for the solution for my own understanding (I'm sure I already got it wrong). My instinct is that this is an arctan integral in the form of 1 / 1 + u^2. Unfortunately, after playing around with it for about 30 minutes I was unable to find a way to get the denominator into the form 1 + u^2.

Am I missing something or did I misjudge the solution?

Anyw help would be greatly appreciated!
 
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What if you use \sqrt{x} as u?

What does du/dx become, and then the rest of the equation?
 
u = root(x) so du = 1/(2*root(x)).

INT [1 / ( 2 * root(x))] * [1 / (1 + x)] dx
= INT [1 / (1 + x)] du = ln|1 + x|

That doesn't seem to work, though. If I differentiate this function I only get the original function with respect to u, not with respect to x since the value of u is eliminated by the value of du during the substitution. I've tried playing with this but I'm not seeing a way to successfully balance the value of du and retain u in the function.

Am I missing the obvious here?
 
You need to express ∫[1 / (1 + x)] du in terms of u with no x at all since you have to integrate with respect to u. So using the substitution u = √x → x = u2, then that gives you ∫du/(1 + u2) which you can now integrate correctly. :smile:
 
Wow, I can't believe I didn't see that. During the exam I tried using root(x) for u but I never thought about root(x)^2 = x. Tunnel vision must have set in- I should have seen that!

So the answer would be arctan(root(x))- I suppose I can console myself knowing I saw arctan(x) correctly. Devil's in the details, unfortunately.

I really appreciate the feedback on this. At least I know what I did wrong now!
 
That's it. Knowing (or figuring out) what to make u in these cases can be a real pain in the neck. And in some cases - such as this - it may not be particularly obvious whether simple substitution or integration by parts is necessary.

I was out last night, so thanks to Bohrok for explaining it more thoroughly.
 
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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