$ \int \sqrt{x^2+81} dx $
Integration by parts formula
\int u(x)v^{'}(x) dx = u(x)v(x) - \int v(x)u^{'}(x) dx
Therefore assume that v^{'}(x) = 1 in this case
Denote the integral
$ I = \int \sqrt{x^2+81} dx$
$ \begin{align*} I &= x \sqrt{x^2+81} - \int \frac{x^2}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - I + 81 \int \frac{1}{\sqrt{x^2+81}} dx \end{align*} $
Therefore
$ 2I = x \sqrt{x^2+81} +81 \int \frac{1}{\sqrt{x^2+81}} dx$
the rest you should do it yourself.
I was told that we are not supposed to give complete solution for homework questions.
IMPROVED EXPLANATION: (BY REQUEST)
$u(x) = \sqrt{x^2+81}$ and
$v(x) = x$, therefore v^{'}(x) = 1
u^{'}(x) = \frac{1}{2}\left(x^2+81\right)^{-\frac{1}{2}} \times 2x = \frac{x}{\sqrt{x^2+81}}
\begin{align*} \int \sqrt{x^2+81} dx &= \int u(x) v^{'}(x) \\ &= u(x)v(x) - \int u^{'}(x) v(x) dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \sqrt{x^2+81} \hspace{3pt} dx + \int \frac{81}{\sqrt{x^2+81}} dx\\ \end{align*}