My Math Forum Integral problem

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 December 9th, 2007, 07:34 AM #1 Newbie   Joined: Dec 2007 Posts: 2 Thanks: 0 Integral problem Integrate on both sides: dy = 1/(c*x^2 + 2*b*c*x) * dx b and c are constants c is positive
 December 25th, 2009, 10:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1752 Use partial fractions.
 December 25th, 2009, 02:52 PM #3 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: Integral problem Hi Polybius; Yes, partial fractions is the way to go, but it doesn't look easy.
December 25th, 2009, 04:02 PM   #4
Math Team

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Re: Integral problem

$\text{Hello, Polybius!}$

$\text{I completed the square . . . then used a trig substitution.}$

Quote:
 $\text{Integrate on both sides: }dy \;=\; \frac{dx}{cx^2 \,+\, 2bcx}$

$\text{We have: }\;dy \;=\;\frac{1}{c}\,\left(\frac{dx}{x^2\,+\,2bx}\rig ht) \;=\;\frac{1}{c}\,\left(\frac{dx}{x^2\,+\,2bx\,+\, b^2\,-\,b^2}\right) \;=\;\frac{1}{c}\,\left(\frac{dx}{(x\,+\,b)^2\,-\,b^2}\right)$

$\text{Integrate: }\;y \;=\;\frac{1}{c}\int\frac{dx}{(x\,+\,b)^2-b^2}$

$\text{Let: }\:x+b \:=\:b\,\sec\theta \qquad\Rightarrow\qquad dx \:=\:b\,\sec\theta\tan\theta\,d\theta$

$\text{Substitute: }\;y \;=\;\frac{1}{c}\int\frac{b\,\sec\theta\tan\theta\ ,d\theta}{b^2\tan^2\theta} \;=\;\frac{1}{bc}\int\frac{\sec\theta}{\tan\theta} \,d\theta \;=\;\frac{1}{bc}\int\frac{\frac{1}{\cos\theta}}{\ frac{\sin\theta}{\cos\theta}}\,d\theta \;=\;\frac{1}{bc}\int\frac{d\theta}{\sin\theta} \;=\;\frac{1}{bc}\int \csc\theta\,d\theta$

$\text{Hence. we have: }\;y \;=\;\frac{1}{bc}\,\ln|\,\csc\theta\,-\,\cot\theta\,| + C$

$\text{Back-substitute: }\:\sec\theta \,=\,\frac{x\,+\,b}{b} \qquad\Rightarrow\qquad \csc\theta \,=\,\frac{x\,+\,b}{\sqrt{x^2\,+\,2bx}}\qquad\Righ tarrow\qquad \cot\theta \,=\,\frac{b}{\sqrt{x^2\,+\,2bx}$

$\text{Therefore: }\;y \;=\;\frac{1}{bc}\,\ln\left|\frac{x\,+\,b}{\sqrt{x ^2\,+\,2bx}}\,-\,\frac{b}{\sqrt{x^2\,+\,2bx}}\right|\,+\,C \;=\;\frac{1}{bc}\,\ln\left|\frac{x}{\sqrt{x^2\,+\ ,2bx}}\right| + C$

 December 25th, 2009, 06:42 PM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Integral problem Partial fractions method gives $\frac{1}{cx^2+2bcx}= \frac{1}{2bc}(\frac1x-\frac1{x+2b})$ which leads to the same result
 December 26th, 2009, 05:32 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1752 Not quite. If b is non-zero, \begin{align*}\int\frac{1}{cx^2\,+\,2bcx}\,dx\,&=\ int\frac{1}{2bc}\left(\frac1x\,-\,\frac{1}{x\,+\,2b}\right)\,dx \\ &=\,\frac{1}{2bc}(\ln|x|\,-\,\ln|x\,+\,2b|)\,+\,\text{C} \\ &=\,\frac{1}{2bc}\ln|\frac{x}{x\,+\,2b}|\,+\,\text {C} \\ &=\,\frac{1}{bc}\ln\frac{|x|}{\sqrt{|x^2\,+\,2bx|} }\,+\,\text{C,}\end{align*} which is subtly different from soroban's result. If b = 0, the answer is -1/(cx) + C.

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