# $\int \sqrt{a^2+x^2} \,dx$ when $\int \sqrt{1+x^2} \, dx$ is known

I messed up calculating $\int\sqrt{a^2+x^2} \, dx$ when $\int \sqrt{1+x^2} \, dx$ is known. What I did

$$\int \sqrt{a^2+x^2} \,dx = \int \sqrt{a^2(1+(x/a)^2)} \, dx = a \int \sqrt{1+(x/a)^2} \,dx$$

now set $u=(x/a)$ and $a\,du=dx$ and use the known formula for $\int \sqrt{1+u^2} \, du$

$$\int \sqrt{a^2+x^2} \,dx= a^2 · (1/2) \left( u \sqrt{1+u^2} + \log(u+\sqrt{1+u^2})\right)$$

now switch back $u=(x/a)$

\begin{align} & a^2 \cdot (1/2) \left( u \sqrt{1+u^2} + \log(u+\sqrt{1+u^2})\right) \\[8pt] & = a^2 \cdot (1/2) \left( (x/a) \sqrt{1+(x/a)^2} + \log((x/a)+\sqrt{1+(x/a)^2})\right) \\[8pt] & = (ax/2) \sqrt{(1/a^2)(a^2+x^2)} + (a^2/2)\cdot\log((x/a)+\sqrt{(1/a^2)(a^2+x^2)} \\ & = (x/2) \sqrt{a^2+x^2} + (a^2/2)\cdot\log\left(1/a(x+\sqrt{a^2+x^2}) \right)\end{align}

$$= (x/2) \sqrt{a^2+x^2}+a^2/2·\log(x+\sqrt{a^2+x^2}) -a$$

Can someone help where's the mistake? I can't find it..

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Remember when you change variables you must change $dx$ to $du$. If $u=\frac{x}{a}$, $\frac{du}{dx}=\frac{1}{a} \implies a\cdot du= dx$
Here is a correct derivation. Let $I(x)=\int \sqrt{1+x^{2}}dx$. Then, your new integral is $$\int \sqrt{a^{2}+x^{2}}dx=\int a \sqrt{1+ \left( \frac{x}{a} \right)^2}dx=a\int a\sqrt{1+u^2}du=a^{2}I(u)=a^{2}I\left(\frac{x}{a}\right)$$ Since $I(x)=\frac{1}{2}(x\sqrt{x^2+1}+\ln(x+\sqrt{1+x^{2}})),$ $$a^{2}I\left(\frac{x}{a}\right)=\frac{1}{2}a^{2}\left(\frac{x}{a}\sqrt{\left(\frac{x}{a}\right)^{2}+1}+\ln\left(\frac{x}{a}+\sqrt{1+\left(\frac{x}{a}\right)^{2}}\right)\right)$$ $$=\frac{1}{2}x\sqrt{x^{2}+a^{2}}+\frac{1}{2}a^{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)-\frac{1}{2}a^{3}$$ With an arbitrary constant of integration, since this is an indefinite integral. Notice that since this constant is arbitrary, we may as well 'absorb' the $-\frac{1}{2}a^{3}$ into it. Hence, your integral is $$=\frac{1}{2}x\sqrt{x^{2}+a^{2}}+\frac{1}{2}a^{2}\ln\left(x+\sqrt{a^{2}+x^{2}}\right)+C$$
Oh, that's a shame. Thanks! Ok, so I've got the first summand right now. But about the second one I'm not sure, it has $a^2/2$ as factor and then I change $u=x/a$ back again. Then inside the log can factor out $1/a$ but it doesn't become the desired solution.. –  fritz Jan 30 '13 at 19:53
So I would get a factor $1/a$ inside the log, which yields in a summand $-a$ whis isn't in the finale solution.. everythin else is ok now.. –  fritz Jan 30 '13 at 20:00
Well, yes, i forgot about the $du$. Now I got the first two summands correctly but from inside the log I got another summand $-a$ which somehow shouldn't be there.. –  fritz Jan 30 '13 at 20:07
Your $-a$ is correct, but I think the answer you have may be the same. See my edit for details. –  Daniel Littlewood Jan 30 '13 at 21:01