<s>[INST] Which formula do I use to integrate $ \int {\sqrt{x^2 + 81} \over 2} \,dx $ I am having trouble with a question really need help please.
$$
\int {\sqrt{x^2 + 81} \over 2} \,dx
$$
I thought about taking the square root off and turning the question into $\frac 12 \int (x^2 +81)^{1/2}\, dx$ but then wondered if I could use the quotient rule.
 [/INST] $$ \int \sqrt{x^2+81} dx $$ 
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*}
$$
</s>