Integration of an inverse function

iceman_ch
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Homework Statement



\int\frac{4}{x(x+3)}

Homework Equations





The Attempt at a Solution



I can get to s certain point and I know I need to do substitution but, everytime I try a substitution it just creates a more difficult problem.

4\int(x^{-1}(x+3)^{-1})

I've tried substitution x^-1 for U and using (x+3)^-1 for dv but, none of it works. If someone could give me a gentle nudge it would be appreciated.

Thanks
 
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This problem is an ideal candidate for the method of "partial fractions".

Try decomposing \frac{1}{x(x+3)} into the form \frac{A}{x}+\frac{B}{x+3} where A and B are constants you need to determine.
 
OH man so obvious. Your the man thank you so much. I havn't had a math class in over a year and now I'm taking diff eq. Bad idea you should definately keep them all together.
 
And, when talking about functions, be careful to distinguish between "reciprocal" and "inverse" functions!
 
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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