Francisca Rodden

2021-12-28

Can trigonometric substitution be used to solve this integral?
The integral in question is
$\int \frac{x+1}{9{x}^{2}+6x+5}dx$
I first completed the square in the denominator giving ${\left(3x+1\right)}^{2}+4$ and proceeded to perform a u-substitution with u=3x+1, du=3 dx, and $x=\frac{u-1}{3}$. After simplifying, I was left with
$\frac{1}{9}\int \frac{u+2}{{u}^{2}+4}du$
It is at this point I used trigonometric substitution with $u=2\mathrm{tan}\theta$ and $du=2\mathrm{sec}2\theta d\theta$ (I'm aware the integral can be written as $\frac{19}{\int }\frac{u}{{u}^{2}+4}du+\frac{19}{\int }\frac{2}{{u}^{2}+4}du$ and solved this way). After performing the trig substitution, I was left with
$\frac{19}{\int }\left(\mathrm{tan}\theta +1\right)d\theta =\frac{19}{\mathrm{ln}}|\mathrm{sec}\theta |+\frac{19}{\theta }+C$
$\frac{19}{\mathrm{ln}\left(\frac{{\left(3x+1\right)}^{2}+4}{2}\right)}+{\frac{19}{\mathrm{tan}}}^{-1}\left(\frac{12}{3x+1}\right)+C$
which is incorrect. The correct answer is
$\frac{1}{18}\mathrm{ln}\left(9{x}^{2}+6x+5\right)+{\frac{19}{\mathrm{tan}}}^{-1}\left(\frac{12}{3x+1}\right)+C$
What went wrong with my trig substitution?

Paineow

Since $u=2\mathrm{tan}\theta$, you have
${u}^{2}=4{\mathrm{tan}}^{2}\theta =4\left({\mathrm{sec}}^{2}\theta -1\right)$
So,
$\mathrm{sec}\theta =\sqrt{\frac{{u}^{2}+4}{4}}$
and therefore
$\frac{19}{\mathrm{ln}\left(\mathrm{sec}\theta \right)}=\frac{1}{18}\mathrm{ln}\left(\frac{{u}^{2}+4}{4}\right)$

alexandrebaud43

There is a simpler way to determine this integral: rewrite first the numerator:
$x+1=\frac{1}{18}\left(18x+6\right)+\frac{23}{}$
split the integral in two:
$\int \frac{x+1}{9{x}^{2}+6x+5}dx=\frac{1}{18}\int \frac{18x+6}{9{x}^{2}+6x+5}dx+\frac{23}{\int }\frac{dx}{{\left(3x+1\right)}^{2}+4}$
$=\frac{1}{18}\mathrm{ln}\left(9{x}^{2}+6x+5\right)+\frac{23}{\int }\frac{dx}{{\left(3x+1\right)}^{2}+4}$
$=\frac{1}{18}\mathrm{ln}\left(9{x}^{2}+6x+5\right)+\frac{29}{\int }\frac{d\left(3x+1\right)}{{\left(3x+1\right)}^{2}+4}$
and use the standard formula
$\int \frac{dx}{{x}^{2}+{a}^{2}}=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)$

nick1337

$⇒\mathrm{ln}|\mathrm{sec}\theta |=\mathrm{ln}\left(\sqrt{{\mathrm{sec}}^{2}\theta }\right)=\frac{1}{2}\mathrm{ln}{\mathrm{sec}}^{2}\theta =\frac{1}{2}\mathrm{ln}\left(1+{\mathrm{tan}}^{2}\theta \right)$
Which for the above:
$\begin{array}{}\frac{1}{9}\mathrm{ln}|\mathrm{sec}\theta |+\frac{1}{9}\theta +{C}_{0}=\frac{1}{9}×\frac{1}{2}\mathrm{ln}\left(1+{\mathrm{tan}}^{2}\theta \right)+\frac{1}{9}\theta +{C}_{0}\\ =\frac{1}{9}×\frac{1}{2}\mathrm{ln}\left(1+\left(\frac{3x+1}{2}{\right)}^{2}\right)+{\mathrm{tan}}^{-1}\left(\frac{3x+1}{2}\right)+{C}_{0}\\ =\frac{1}{18}\mathrm{ln}\left(9{x}^{2}+6x+5\right)+{\mathrm{tan}}^{-1}\left(\frac{3x+1}{2}\right)+\left({C}_{0}-\frac{1}{18}\mathrm{ln}\left(4\right)\right)\\ =\frac{1}{18}\mathrm{ln}\left(9{x}^{2}+6x+5\right)+{\mathrm{tan}}^{-1}\left(\frac{3x+1}{2}\right)+C\end{array}$

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