MHB S6.7.r.19 Rational Expression Integral (complete the square?)

[karush]

$\Large{S6.7.R.19}$

$$\displaystyle
I=\int\frac{x+1}{9{x}^{2}+6x+5}\, dx
=\frac{1}{18}\ln\left({9{x}^{2}+6x+5}\right)
+\frac{1}{9}\arctan\left[{\frac{1}{2}\left(3x+1\right)}\right]+C
$$
$\text{from the given I thought completing the square would be the way to solve this} \\$
$\text{but I don't see how this would result in the answer.}$

 

[Prove It]

$\Large{S6.7.R.19}$

$$\displaystyle
I=\int\frac{x+1}{9{x}^{2}+6x+5}\, dx
=\frac{1}{18}\ln\left({9{x}^{2}+6x+5}\right)
+\frac{1}{9}\arctan\left[{\frac{1}{2}\left(3x+1\right)}\right]+C
$$
$\text{from the given I thought completing the square would be the way to solve this} \\$
$\text{but I don't see how this would result in the answer.}$

I would start by rewriting it in two separate fractions, one where a u substitution can be used... Notice that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( 9\,x^2 + 6\,x + 5 \right) = 18\,x + 6 \end{align*}$, so we should probably rewrite the integral as...

$\displaystyle \begin{align*} \int{ \frac{x + 1}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } &= \frac{1}{18} \int{ \frac{18\,x + 18}{9\,x^2 + 6\,x + 5} } \\ &= \frac{1}{18} \int{ \frac{18\,x + 6}{9\,x^2 + 6\,x + 5} + \frac{12}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } \end{align*}$

The first term can be integrated with substitution $\displaystyle \begin{align*} u = 9\,x^2 + 6\,x + 5 \implies \mathrm{d}x = \left( 18\,x + 6 \right) \,\mathrm{d}x \end{align*}$ and the second can be solved with your idea of completing the square in the denominator.

 

[karush]

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( 9\,x^2 + 6\,x + 5 \right) = 18\,x + 6 \end{align*}$

$\displaystyle \begin{align*} \int{ \frac{x + 1}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } &= \frac{1}{18} \int{ \frac{18\,x + 18}{9\,x^2 + 6\,x + 5} } \\ &= \frac{1}{18} \int{ \frac{18\,x + 6}{9\,x^2 + 6\,x + 5} + \frac{12}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } \end{align*}$

$\displaystyle \begin{align*} u = 9\,x^2 + 6\,x + 5 \implies \mathrm{d}x = \left( 18\,x + 6 \right) \,\mathrm{d}x \end{align*}$

$\displaystyle
\frac{1}{18} \int{\frac{18x + 6}{9\,x^2 + 6x + 5}} \, dx
= \frac{1}{18}\int\frac{1}{u}\, du =\frac{1}{18}\ln\left({9\,x^2 + 6x+5}\right)$

 

[Prove It]

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( 9\,x^2 + 6\,x + 5 \right) = 18\,x + 6 \end{align*}$

$\displaystyle \begin{align*} \int{ \frac{x + 1}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } &= \frac{1}{18} \int{ \frac{18\,x + 18}{9\,x^2 + 6\,x + 5} } \\ &= \frac{1}{18} \int{ \frac{18\,x + 6}{9\,x^2 + 6\,x + 5} + \frac{12}{9\,x^2 + 6\,x + 5} \,\mathrm{d}x } \end{align*}$

$\displaystyle \begin{align*} u = 9\,x^2 + 6\,x + 5 \implies \mathrm{d}x = \left( 18\,x + 6 \right) \,\mathrm{d}x \end{align*}$

$\displaystyle
\frac{1}{18} \int{\frac{18x + 6}{9\,x^2 + 6x + 5}} \, dx
= \frac{1}{18}\int\frac{1}{u}\, du =\frac{1}{18}\ln\left({9\,x^2 + 6x+5}\right)$

That is correct so far. Notice you can leave out the absolute value signs in the logarithm as the quantity is always nonnegative anyway.

Now how about the second part?