MHB Q5|8.5.17 int of rational expression...

[karush]

$\tiny{Q5|8.5.17}$
$\textsf{Evaluate}$
\begin{align*}\displaystyle
I&=\int_0^{9}\frac{x^3 \, dx}{x^2+18x+81}
\color{red}{=243\ln2-162}
\end{align*}

OK even before I try to get this answer
trying to see what road to take
u substituion
long division and remainder
partial fractions (the denominator is a square $(x+9)^2$)
other?

I tried $u=x+9$ $du=dx$ but it got ? fast

 

[MarkFL]

Let's try your substitution of:

$$u=x+9\implies du=dx$$

$$I=\int_9^{18}\frac{(u-9)^3}{u^2}\,du$$

By the binomial theorem:

$$(u-9)^3=u^3+3u^2(-9)+3u(-9)^2+(-9)^3=u^3-27u^2+243u-729$$

Hence:

$$I=\int_9^{18} u-27+243u^{-1}-729u^{-2}\,du$$

Now, you can integrate term by term. :D