$$\begin{align*}\int\sqrt{x^2+a^2}dx&=x\sqrt{x^2+a^2}-\int xd(\sqrt{x^2+a^2})\\&=x\sqrt{x^2+a^2}-\int\frac{x^2}{\sqrt{x^2+a^2}}dx\\&=x\sqrt{x^2+a^2}-\int\frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}dx\\&=x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}dx + a^2\int\frac{1}{\sqrt{x^2+a^2}}dx\end{align*}$$

By the result of # 4626, we know $$\int\frac{1}{\sqrt{x^2+a^2}}dx=\ln|x+\sqrt{x^2+a^2}|+C$$

Setting this result back to the previous relation can lead to the final answer

$$\int\sqrt{x^2+a^2}dx=\boxed{\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln{|x+\sqrt{x^2+a^2}|}+C}$$