I want to show that this equation holds for positive a. $$\int \sqrt{a^2 + u^2}\;du = \frac{u}{2} \sqrt{a^2 + u^2}+ \frac{a^2}{2}\;ln(u+\sqrt{a^2+u^2}))+ C $$ My attempt is this: $$u=a.tan(t)\\du=a(1+tan^2(t))dt\\\int \sqrt{a^2 + u^2}\;du=\int \sqrt{a^2 + a^2tan^2(t)}\;du =\int |a.sec(t)|.a.(sec^2(t))\;dt=a^2\int |sec^3(t)|dt$$
And we know that: $$\int sec^3(t)dt=\frac{1}{2}(tan(t).sec(t)+ln|sec(t)+tan(t)|)=I$$ Then I made an attempt to rewrite the right side of my equation. $$\frac{u}{2} \sqrt{a^2 + u^2}= \frac{a.tan(t)}{2}.|a.sec(t)|=\frac{a^2}{2}tan(t)|sec(t)|=II\\\frac{a^2}{2}\;ln(u+\sqrt{a^2+u^2})=\frac{a^2}{2}.ln(a.tan(t)+|a.sec(t)|)=\frac{a^2}{2}ln(a(tan(t)+|sec(t)|))=III$$ If we multiply I by a^2: $$I=\frac{a^2}{2}(tan(t).sec(t))\;[which\;looks\;similar\;to\;II]\\+\frac{a^2}{2}ln|sec(t)+tan(t)|\;[which\;looks\;similar\;to\;III] $$ I'm confused what should I do from now on; My first problem is with absolute values and my second is the fact that III has an "a" element inside the parenthesis of ln but I doesn't. Any help would be appreciated!
A natural substitution for this integral is $u=a\sinh t$. Then,
$$\int \sqrt{a^2 + u^2}\;du =a^2\int \sqrt{1+\sinh^2t}\cosh tdt=a^2 \int \cosh^2tdt$$
$$=\frac12 a^2\int (1+\cosh 2t)dt=\frac 12 a^2 (t + \frac12\sinh 2t)+C$$
$$=\frac 12 a^2 \sinh^{-1} \frac ua + \frac 12a^2\sinh t \cosh t+C$$ $$= \frac{a^2}{2}\;\ln(u+\sqrt{a^2+u^2})+ \frac{1}{2}u \sqrt{a^2 + u^2}+C $$
where $\sinh^{-1} x= \ln(x+\sqrt{1+x^2})$ is used.
Edit:
The use of substitution $u=a\tan t$ could be somewhat cumbersome. Continue with
$$I=\int \sec^3t dt= \int (1+\tan^2t)\sec tdt = \int \sec tdt + \int \tan t d(\sec t)$$ $$= \ln(\sec t + \tan t) + \tan t \sec t - I$$
Then,
$$I = \frac12 \ln(\sec t + \tan t) + \frac12 \tan t \sec t + C$$
Then, plug back $\tan t = \frac ua$ and $\sec t = \frac{\sqrt{a^2+u^2}}a$ to recover the given result,
$$a^2\int \sec^3t dt= \frac12a^2 \ln\frac{\sqrt{a^2+u^2}+u}a + \frac12 u\sqrt{a^2+u^2}+ C$$