indefinite integral $$\int\sqrt {x^2 + a^2} dx$$
After some transformations and different substitution, I got stuck at this
$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$
I am not sure I am getting the first step correct. Tried substituting $ x=a\tan \theta$ but that doesn't help either.
Consider the integral \begin{align} I = \int \sqrt{x^{2}+ a^{2}} \, dx. \end{align} Make the substitution $x = a \sinh(t)$, $dx = a \cosh(t) dt$, it is seen that \begin{align} I &= a \int \sqrt{ a^{2} (1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\ &= a^{2} \int \sqrt{\cosh^{2}(t)} \cosh(t) \, dt \\ &= a^{2} \int \cosh^{2}(t) \, dt \\ &= \frac{a^{2}}{2} \int (1 + \cosh(2t)) dt \\ &= \frac{a^{2}}{2} \left[ t + \frac{1}{2} \sinh(2t) \right] \\ &= \frac{a^{2}}{2} \left[ t + \sinh(t) \cosh(t) \right]. \end{align} Now back substitute to obtain \begin{align} \int \sqrt{x^{2}+ a^{2}} \, dx &= \frac{a^{2}}{2} \left[ \sinh^{-1}(x/a) + (x/a) \cosh(\sinh^{-1}(x/a)) \right] \\ &= \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \sinh^{-1}\left( \frac{x}{a} \right). \end{align}
(I assume $a>0$, which is not restrictive.)
This can be treated in a way very similar to $\int\sqrt{a^2-x^2}\,dx$. Set $x=a\sinh t$, so $dx=a\cosh t\,dt$ and $$ \sqrt{a^2\sinh^2t+a^2}=a\sqrt{\sinh^2t+1}=a\cosh t $$ Thus you have to compute $$ \int\cosh^2t\,dt $$ The fundamental relation is $$ \cosh^2t-\sinh^2t=1 $$ so $$ \int\cosh^2t\,dt-\int\sinh^2t\,dt=t\rlap{\qquad(*)} $$ (I'll omit the constant of integration). But, integrating by parts, $$ \int\sinh t\sinh t\,dt=\cosh t\sinh t-\int\cosh t\cosh t\,dt $$ or $$ \int\sinh^2 t\,dt=\cosh t\sinh t-\int\cosh^2t\,dt $$ and, replacing in $(*)$ we get $$ 2\int\cosh^2t\,dt=t-\cosh t\sinh t $$ Now it's just a problem of back substitution: $\sinh t=\frac{x}{a}$ and $$ \cosh t=\frac{\sqrt{x^2+a^2}}{a} $$ while, from $$ a\frac{e^t-e^{-t}}{2}=x $$ we get $$ ae^2t-2xe^t-a=0 $$ or $$ e^t=\frac{x+\sqrt{x^2+a^2}}{a} $$ and so $$ t=\log\bigl(x+\sqrt{x^2+a^2}\,\bigr)-\log a $$
Let $x=a\tan\theta$, so $dx=a\sec^{2}\theta d\theta$ to get
$\int\sqrt{x^2+a^2}\;dx=a^2\int\sec^{3}\theta\;d\theta$. Using integration by parts with $u=\sec\theta$ and $dv=\sec^{2}\theta\;d\theta$ gives
$\hspace{.6 in}\sec^{3}\theta\;d\theta=\frac{1}{2}\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C$,
so $\int\sqrt{x^2+a^2}=\frac{a^2}{2}\left(\frac{\sqrt{x^2+a^2}}{a}\frac{x}{a}+\ln\left|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}\right|\right)+C$
$\hspace{.9 in}=\frac{1}{2}\left({x\sqrt{x^2+a^2}}+a^2\ln\left(\sqrt{x^2+a^2}+x\right)\right)+C$
Here we have another way to see this: $$ \int \sqrt{x^2+a^2} dx $$ using the substitution $$ t=x+\sqrt{x^2+a^2}\\ \sqrt{x^2+a^2}=t-x $$ and squaring we have $$ a^2 =t^2-2tx\\ x=\frac{t^2-a^2}{2t}. $$ Finally we can use: $$ dx=\frac{2t(t)-(t^2-a^2)(1)}{2t^2}dt = \frac{t^2+a^2}{2t^2}dt\\ \sqrt{x^2+a^2}=t-\frac{t^2-a^2}{2t}=\frac{t^2+a^2}{2t}. $$ Thus: $$ \int \sqrt{x^2+a^2} dx = \int \frac{t^2+a^2}{2t} \frac{t^2+a^2}{2t^2}dt=\int \frac{(t^2+a^2)^2}{4t^3}dt $$ which is elementary, if we expand the square of the binomial: $$ \int\frac{t}{4}dt+\int\frac{a^2}{2t}dt+\int\frac{a^4}{4t^3}dt=\frac{t^2}{8}+a^2\ln\sqrt{t}-\frac{a^4}{16t^4}, $$ where as stated $t=x+\sqrt{x^2+a^2}.$
$$\sqrt{x^2+a^2}dx = a^2\sqrt{\Big(\frac{x}{a}\Big)^2+1}\,d\frac{x}{a}.$$ Set $\sinh\theta=\frac{x}{a}$. $$\sqrt{1+\sinh^2\theta}d\sinh\theta=\cosh^2\theta d\theta=\frac{1}{2}(\cosh2\theta+1)d\theta=\frac{1}{4}d(\sinh2\theta+2\theta).$$
$\displaystyle \int\sqrt{x^{2}+a^{2}}dx$
$\displaystyle x=a.\ sh(t)\Rightarrow dx=a\ ch(t)dt$
$\displaystyle \int\sqrt{x^{2}+a^{2}}dx =a^{2}\int\ ch^{2}(t)dt=\frac{a^{2}}{2}\int(1+\ ch(2t))dt $
$\displaystyle =\frac{a^{2}}{2}(t+\frac{\ sh(2t)}{2})+c$
$\displaystyle t=\ sh^{-1}(\frac{x}{a})=ln(\frac{x}{a}+\sqrt{1+\frac{x^{2}}{a^{2}}})=ln(x+\sqrt{x^{2}+a^{2}})-ln(a)$
$\displaystyle\ sh(2t)=2\ sh(t)\ ch(t)=2\frac{x}{a}\sqrt{1+\frac{x^{2}}{a^{2}}}=2\frac{x}{a^{2}}\sqrt{a^{2}+x^{2}}$
$\displaystyle \int\sqrt{x^{2}+a^{2}}dx=\frac{a^{2}}{2}ln(x+\sqrt{x^{2}+a^{2}})+\frac{x}{2}\sqrt{x^{2}+a^{2}}+C$
