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

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  • $\begingroup$ Everything is fine until your end substitution. The inside of your log is $(\sec^2\theta) /2$ not $\sec\theta$ $\endgroup$ Commented Oct 6, 2021 at 9:17
  • $\begingroup$ Thank you for the quick reply! I'm a bit confused. Why would the argument of the log be $(\sec^2\theta)/2$? The antiderivative of $\tan\theta$ is $\ln|\sec\theta|+C$. $\endgroup$ Commented Oct 6, 2021 at 9:28
  • $\begingroup$ What I'm saying is you incorrectly backsubstituted and that's what's inside now when it should just be $\sec\theta$ $\endgroup$ Commented Oct 6, 2021 at 9:58
  • $\begingroup$ Thank you very much @NinadMunshi; I appreciate the help! Sometimes it's easy to forget those little important details. $\endgroup$ Commented Oct 6, 2021 at 23:44

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Since $u=2\tan\theta$, you have$$u^2=4\tan^2\theta=4(\sec^2\theta-1).$$So,$$\sec\theta=\sqrt{\frac{u^2+4}4}$$and therefore$$\frac19\ln(\sec\theta)=\frac1{18}\ln\left(\frac{u^2+4}4\right).$$

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  • $\begingroup$ Thank you for the quick reply! After re-substituting $u = 3x+1$, I'm confused as to why there is a factor of $1/4$ in the argument of the logarithm. $\endgroup$ Commented Oct 6, 2021 at 9:26
  • $\begingroup$ Why not? If $u=3x+1$, then$$\frac{u^2+4}4=\frac{9x^2+6x+5}4.$$But note that$$\ln\left(\frac{9x^2+6x+5}4\right)=\ln(9x^2+6x+5)-\ln4.$$ $\endgroup$ Commented Oct 6, 2021 at 9:43
  • $\begingroup$ Thank you very much! I just recognized that I can rewrite the solution with a new constant $C_1 := C - \ln 4/18$. $\endgroup$ Commented Oct 6, 2021 at 9:50
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There is a simpler way to determine this integral: rewrite first the numerator: $$x+1=\frac 1{18}(18x+6)+\frac 23,$$ split the integral in two: \begin{align} \int \frac{x+1}{9x^2+6x+5}\,\mathrm dx&=\frac 1{18}\int \frac{18x+6}{9x^2+6x+5}\,\mathrm dx+\frac23\int\frac{\mathrm dx}{(3x+1)^2+4}\\ &=\frac 1{18}\ln(9x^2+6x+5)+\frac23\int\frac{\mathrm dx}{(3x+1)^2+4}\\ &=\frac 1{18}\ln(9x^2+6x+5)+\frac29\int\frac{\mathrm d(3x+1)}{(3x+1)^2+4}, \end{align} and use the standard formula $$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\,\arctan \Bigl(\frac xa\Bigr).$$

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    $\begingroup$ @DarshanPatil: Thanks for the edit! I definitely should in detail what I typed before posting… $\endgroup$
    – Bernard
    Commented Oct 6, 2021 at 11:14
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$$\begin{align*} \int \frac{x+1}{9x^2+6x+5}dx &=\int\frac{x+1}{(3x+1)^2+2^2}dx\\ (u = 3x+1)\\ & = \frac{1}{9}\int\frac{u+2}{u^2+2^2}du\\ (u = 2\tan\theta)\\ &=\frac{1}{9}\int\frac{(2\tan\theta+2)(2\sec^2\theta)}{2^2(1+\tan^2\theta)}d\theta\\ &=\frac{1}{9}\int(\tan\theta+1)d\theta\\ & = \frac{1}{9}\ln{|\sec\theta|} + \frac{1}{9}\theta + C_0 \end{align*}$$

Now, $|x| = \sqrt{x^2}$

$\implies \ln|\sec\theta| = \ln{(\sqrt{\sec^2{\theta}})} = \frac{1}{2}\ln{\sec^2{\theta}} = \frac{1}{2}\ln{(1+\tan^2{\theta})}$

Which for the above:

$$\begin{align*} \frac{1}{9}\ln{|\sec\theta|} + \frac{1}{9}\theta + C_0 & = \frac{1}{9}\times\frac{1}{2}\ln{(1+\tan^2{\theta})}+ \frac{1}{9}\theta + C_0\\ & = \frac{1}{9}\times\frac{1}{2}\ln\left(1+\left(\frac{3x+1}{2}\right)^2\right) + \tan^{-1}\left(\frac{3x+1}{2}\right)+C_0\\ & = \frac 1{18}\ln(9x^2 + 6x +5) + \tan^{-1}\left(\frac{3x+1}{2}\right)+(C_0 - \frac 1{18}\ln(4))\\ & = \frac 1{18}\ln(9x^2 + 6x +5) + \tan^{-1}\left(\frac{3x+1}{2}\right)+C \end{align*}$$

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  • $\begingroup$ Did that take 40 minutes to type! $\endgroup$
    – ACB
    Commented Oct 6, 2021 at 10:07

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