Mastering Trig Substitution for Evaluating Integrals of 1/x(4x+1)^1/2 in Calc II

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evaluate the integral of 1/x(4x+1)^1/2

so what i am thinking is that i will have to use a trig substution some how for
4x+1

maybe x=secu since the substution for a^2+x^2=sec^2u

but i am not sure how to go ahead and do it

please help me
 
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No. Don't do trig. Try u^2=4*x+1. You'll get a simple rational function you can then do with partial fractions.
 
GravityGirl said:
evaluate the integral of 1/x(4x+1)^1/2

so what i am thinking is that i will have to use a trig substution some how for
4x+1

maybe x=secu since the substution for a^2+x^2=sec^2u

but i am not sure how to go ahead and do it

please help me
Well, as Dick pointed out, you should let u = \sqrt{4x + 1}
Or, in general, if you meet some radical, then it's common to let u = that whole radical, and then go from there.

You only let x = sec(u), when you meet: \sqrt{x ^ 2 - 1}, or when encountering \sqrt{x ^ 2 - a ^ 2}, you then let x = a sec(u).

Can you go from here? :)
 
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I guess if she really wanted to, she could have done x= 1/4 tan^2 u. Takes much much longer though.
 
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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