What Is the Best Method to Integrate dx/(x*sqrt(9+16x^2))?

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Homework Statement




indefinite integral: dx/(x*sqrt(9+16x^2))


Homework Equations



Trig. Substitutions or parts??

The Attempt at a Solution



I tried using integration by parts but its got pretty messy...it also resembles a tan trig substitution, but it's within a square root. I'm stumped and can't figure out where to start...

Can anyone help?

Thanks!
 
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Don't use integration by parts. Rewrite the problem as
\int\frac{xdx}{x^2\sqrt{9+ 16x^2}}
and let u= 9+ 16x^2.
 
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Factor a 3 out of \sqrt{9 +16x^2}

\displaystyle \sqrt{9 +16x^2}=\sqrt{9\left(1+\frac{16x^2}{9}\right)}=3 \sqrt{1+\frac{16x^2}{9}}=3 \sqrt{1+\left( \frac{4x}{3} \right)^2 }

We know that 1+tan2(θ) = sec2(θ) , so let 4x/3 = tan(θ), (4/3)dx = sec2(θ)dθ.

(I'm slow at typing, so HallsofIvy responded while I was typing. He usually has better ideas than I do. Good luck!)
 
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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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