Integration by Parts on $\sqrt{1+x^2}dx$ I have been asked to show that $$\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\int\frac{1}{{\sqrt{1+x^2}}}dx$$
I'm aware of how to do this with trig substitution, but the question is specifically regarding integration by parts, and the only examples I can find keep using trig subs.
So far I have integrated twice. The first time was with $u=\sqrt{1+x^2},v'=1$. Resulting in the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\int\frac{x}{\sqrt{1+x^2}}dx$$
The first term is almost correct. Integrating this new integral, with $u=\frac{x}{\sqrt{1+x^2}}, v'=1$. Gave me the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{2}\int\frac{x}{\sqrt{1+x^2}}dx$$
From here I'm pretty much stuck. I can't seem to figure out how I am supposed to get rid of the middle term. The final integral can be put into the right form if I divide by $x$, but then the first term loses it's $x$ scalar and I still can't drop the middle term.
I feel as though I've made a mistake on one of the integrations, but I've been looking over it for that long that the whole thing is blurry.

Alternative method, but not really so different from the other answer:
\begin{align}
\int\sqrt{1+x^2}\,dx
&=
\int\frac{1+x^2}{\sqrt{1+x^2}}\,dx
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+
\int x\frac{x}{\sqrt{1+x^2}}\,dx 
\qquad \biggl(u=x, v'=\frac{x}{\sqrt{1+x^2}}\biggr)
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+x\sqrt{1+x^2}-\int\sqrt{1+x^2}\,dx
\end{align}
Therefore
$$
2\int\sqrt{1+x^2}\,dx
=
x\sqrt{1+x^2}+\int\frac{1}{\sqrt{1+x^2}}\,dx
$$
