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I have been attempting to solve for the integral of the below equation but have had no luck so far. After searching online I have only managed to find one mention of it but with no explanation (the below picture).

I have tried using the quotient rule by parts formula, and substitution but have found myself running circles each time and getting nowhere near the proposed answer on the site. I need this integral for a larger integral I'm doing but need to know its proof so I may properly explain its use in my paper.

I know I'm not really providing much here, but I've become completely lost with this integral. Any help is much appreciated.

http://www.sosmath.com/tables/integral/integ11/img3.gif

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  • $\begingroup$ Hint: $t=x^2+a^2$. $\endgroup$
    – Lucian
    Sep 11, 2015 at 1:42
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    $\begingroup$ Next time, just state the problem in short $\endgroup$
    – Shailesh
    Sep 11, 2015 at 1:51
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    $\begingroup$ The formula in the title subtracts $x^2$ from $a^2$, but the formula in the picture in the question body adds $x^2$ to $a^2$. Which integral do you want? $\endgroup$
    – David K
    Sep 11, 2015 at 2:05
  • $\begingroup$ The proof of the equation in the picture is simply the fact that the derivative of the expression on the right-hand side of the equation is the expression inside the integral on the left-hand side. What do you need other than that to "explain its use" in your paper? $\endgroup$
    – David K
    Sep 11, 2015 at 2:28
  • $\begingroup$ Welcome to Maths.SE! Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. $\endgroup$
    – Lord_Farin
    Sep 13, 2015 at 19:43

3 Answers 3

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Let $u = x^2+a^2$ as has been suggested. Then $\frac{du}{dx} = 2x$ or equivalently $dx = \frac{du}{2x}$. Substituting these quantities into the integral gives

\begin{align} \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + C = \sqrt{x^2+a^2} + C \end{align}

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  • $\begingroup$ Okay this fixes a lot, because I tried using t = a^2 +x^2 but I messed up on the substitution. Silly mistake, but thanks a lot, you and Lucian help a lot Ill accept answer in a sec after i look this over a bit and properly understand what I did wrong. $\endgroup$
    – 23scurtu
    Sep 11, 2015 at 2:19
  • $\begingroup$ @23scurtu No problem, let me know if you don't understand something. $\endgroup$
    – Dipole
    Sep 11, 2015 at 2:21
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You need only the "fundamental theorem of calculus" here!

Note that $$ \int_{x} \frac{x}{\sqrt{x^{2}+a^{2}}} = \int_{x}D(x^{2}+a^{2})^{1/2} = (x^{2}+a^{2})^{1/2} + \text{some constant}. $$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – user642796
    Sep 11, 2015 at 7:01
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Let's use U-substitution to solve this problem!

$$\int \frac{x}{\sqrt{x^{2}+a^{2}}}dx$$

$$\text{Let: }\qquad u = x^2+a^2\quad \text{and}\quad du = 2xdx$$

Therefore, \begin{align} \require{cancel}\\ \int \frac{x}{\sqrt{x^{2}+a^{2}}}dx &= \frac{1}{2}\int\frac{1}{\sqrt u}du\\ &= \frac{1}{\cancel2} (\cancel2\sqrt u) + C\\ &= \sqrt u + C\\ &= \sqrt{x^2+a^2} + C\\ \end{align}

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  • $\begingroup$ when you canceled out the 2, I thought that there was only one 2 that came from the 1/2. where did the second one come from? $\endgroup$
    – 23scurtu
    Sep 11, 2015 at 2:32
  • $\begingroup$ The $\int \frac{1}{\sqrt u}du$ is $2\sqrt u + C$. Do you want me to explicitly have that worked out in the answer? $\endgroup$
    – Slinky
    Sep 11, 2015 at 2:35
  • $\begingroup$ Oh crap! I see how it works. raising the power to the u^-1/2 requires multiplying by 2 to cancel out the 1/2 that would be carried down through the derivation of u^1/2. Adding that would prob help anyone else coming by this question. $\endgroup$
    – 23scurtu
    Sep 11, 2015 at 2:46
  • $\begingroup$ @23scurtu Don't think of it as multiplying by 2 to cancel out the $\frac{1}{2}$ in the long run, just think of the short term integral, $\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}=2u^{\frac{1}{2}} = 2\sqrt u$. $\endgroup$
    – Slinky
    Sep 11, 2015 at 2:52
  • $\begingroup$ alright thankks $\endgroup$
    – 23scurtu
    Sep 11, 2015 at 3:07

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