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So here is my equation:

$$\int{\frac{dx}{(x^2 + d^2)^{1/2}}}$$

Is there any way to solve this? Thanks! Also, $d$ is just a constant.

1 Answers1

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I hope you won't mind if I use $a$ instead of your $d$, which interferes too much visually with the notation for the derivative.

The simplest approach uses hyperbolic functions. Let $x=a\sinh t$. We end up with $\int dt$, which is $t+C$. But $t=\operatorname{arcsinh}(x/a)$.

Hyperbolic functions are often absent from first calculus courses. Then things get a lot messier. The standard substitution is $x=a\tan \theta$. Then $dx=a\sec^2\theta\,d \theta$. We end up having to find $\int \sec\theta\,d\theta$.

Possibly the integral of $\sec\theta$ is part of your standard list of integrals. If it isn't, we need to find it.

Note that $\sec\theta=\frac{1}{\cos\theta}=\frac{\cos\theta}{1-\sin^2\theta}$. Make the substitution $u=\sin\theta$, and we end up with $\int \frac{1}{1-u^2}\,du$.

But $\frac{1}{1-u^2}=\frac{1}{2}\left(\frac{1}{1-u}+\frac{1}{1+u}\right)$. Now we are at a familiar problem.

André Nicolas
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    Partial fractions? Crap I bit off more than I can chew haha. I was just trying to proove a law in physics to see if I fully understood it. Thank you though! – Athan Clark Mar 05 '13 at 04:31
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    In Physics, they are not shy about using hyperbolic functions. – André Nicolas Mar 05 '13 at 04:34
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    Do you know of any good resources for learning about how to work with them, aside from Wikipedia? – Athan Clark Mar 05 '13 at 04:37
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    Actually, Wikipedia is good here. For basic work, there is nothing much that needs to be known about hyperbolic functions apart from the basic identities and formulas. Differentiation is even simpler than with trig functions: the derivative of $\sinh$ is $\cosh$, and the derivative of $\cosh$ is $\sinh$. The pair $(\cosh t,\sinh t)$ parametrizes the rectangular hyperbola $x^2-y^2=1$, just like $(\cos t,\sin t)$ parametrizes the unit circle. Standard "big" calculus books, like Stewart, have a few pages on hyperbolic functions. – André Nicolas Mar 05 '13 at 04:49