Integrate:
$$\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx$$
Here's my attempt, using integration by parts in the first step:
\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\int \:-\frac{x^2}{\left(9+x^2\right)^{1.5}}dx\right]^1_0 \\
&= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\left(\frac{x}{\sqrt{9+x^2}}-\ln \left(\frac{1}{3} \left|x+\sqrt{9+x^2}\right|\right)\right)\right]^1_0 \\
&= \left[\ln \left(\frac{1}{3}\left|x+\sqrt{x^2+9}\right|\right)\right]^1_0 \\
&= \ln \left(\frac{1+\sqrt{10}}{3}\right)
\end{align}
I tried to solve this using substitution and I keep getting stuck at the same place. For the sake of understanding, I turned the definite integral into an indefinite one by remove the upper and lower bounds, since that's quite easy to do. I'm more concerned with the final result without the actual values. Here's my attempt:
\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &\stackrel{{u = \frac{x}{3}}, dx = 3\,du} = \int\frac{3}{\sqrt{9u^2+9}}\,\mathrm{d}u \\
&= \int \dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u \\
&\stackrel{{u=\tan(v)} \to \mathrm{d}u=\sec^2(v)\,\mathrm{d}v} = \int \frac{\sec^2(v)}{\sqrt{\tan^2(v)+1}}\,\mathrm{d}v \\
&= \int \sec(v)\,\mathrm{d}v \\
&= \int \frac{\sec(v) (\tan(v)+\sec(v))}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \int \frac{\sec(v) \tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \cdots
\end{align}
Here is where I am stuck. I think that there should be another substitution here in order to get the final integral that can be solved, but I'm not sure which term to choose. I think it would have to be $w = w=\tan\left(v\right)+\sec\left(v\right)$, but I'm not $100 \%$ sure.
This really caught me off by surprise because it looks rather simple but boy is it tricky to do.
Best Answer
\begin{align} \int\frac{1}{\sqrt{x^2+9}}\,dx &=\int\sec(u)\,du\quad &(x=3\tan u)\\ &=\int \frac{\sec^2(u)+\tan(u)\sec(u)}{\tan(u)+\sec(u)}\,du\\ &=\int\frac{1}{s}\,ds&(s=\tan(u)+\sec(u))\\ &=\log(s)+C\\ &=\log(\tan(u)+\sec(u))+C\\ &=\log\bigg(\frac13(\sqrt{x^2+9}+x)\bigg)+C \end{align}