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How to evaluate $$\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$$ I tried trigonometric substitution $x + a = b \sec \theta$ and I encountered $$\int \frac{\tan \theta}{ (b - a \cos \theta) \sqrt{\tan^2 \theta}}d\theta$$ how to handle this term $\displaystyle \frac{\tan \theta}{|\tan \theta|}$?

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@MichaelHardy yes sorry –  hasExams Jun 5 '13 at 15:43
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The expression you should get might depend on whehter $b<a$ or $b>a$. –  Michael Hardy Jun 5 '13 at 17:17
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2 Answers 2

up vote 3 down vote accepted

I substituted $x=1/y$ and got

$$-\int \frac{dy}{\sqrt{1+2 a y+(a^2-b^2) y^2}}$$

I then completed the square in the square root to get

$$-\frac{1}{\sqrt{a^2-b^2}} \int \frac{dy}{\displaystyle\sqrt{\left(y+\frac{a}{a^2-b^2}\right)^2-\left(\frac{a}{a^2-b^2}\right)^2}}$$

Now let

$$y=\frac{a}{a^2-b^2} (\cosh{u}-1)$$

Then the integral becomes

$$-\frac{1}{\sqrt{a^2-b^2}} \int du \frac{\sinh{u}}{\sinh{u}} = -\frac{1}{\sqrt{a^2-b^2}} u+C$$

where $C$ is a constant of integration. Back substituting, we get

$$\int \frac{dx}{x \sqrt{(x+a)^2-b^2}} = -\frac{1}{\sqrt{a^2-b^2}}\text{arccosh}{\left(\frac{a^2-b^2}{a x}+1 \right)} + C $$

Use $\text{arccosh}{z} = \log{(z+\sqrt{z^2-1})}$ to express the result in terms of logs.

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First, examine separately intervals where $\tan\theta>0$ and where $\tan\theta<0$. After that, see if you can express the answer in a way that's less-than-explicitly piecewise (perhaps by using absolute values).

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let it be $I$ if $\theta>0$, then if $\theta<0$, then my answer is $-I$, but Mathematica gives me something negative of $\log(-x)$ for my original integration $\frac{1}{x \sqrt{(x+a)^2 - b^2}}$ –  hasExams Jun 5 '13 at 15:53
    
mathematica evaluates it to $$ \frac{\log (-x)-\log \left(\sqrt{a^2-b^2} \sqrt{(a+x)^2-b^2}+a (a+x)-b^2\right)}{\sqrt{a^2-b^2}}$$ I can't get this $\log(-x)$ part –  hasExams Jun 5 '13 at 16:16
    
Before checking details, by initial reaction is that $\log|x|$ occurs as the antiderivative of $1/x$, and that's the same as $\log x$ if $x>0$ and it's the same as $\log(-x)$ if $x<0$. –  Michael Hardy Jun 5 '13 at 17:12
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