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I was attempting to solve a more daunting integral than usual as a fun challenge, and two more integrals came out as part of the answer.

$$ \int \frac x {x^2 + \frac 3 4} dx + {\frac 3 2}\int {\frac 1 {x^2 + \frac 3 4}} dx $$

The second integral is pretty clearly $\frac 3 {\sqrt 3} \tan^{-1}(\frac {2x} {\sqrt 3})$. However, I tried to use the same reasoning for the first integral, and figured a substitution for $x^2$ would yield the following:

$$ \frac 1 2 \int {\frac 1 {u^2 + \frac 3 4}} du = \frac {\sqrt 3} 3 \tan^{-1}(\frac {2u} {\sqrt 3}) $$

I didn't simplify further, because it seems I was wrong. The solution I found was to perform the substitution and evaluate the integral as ${\frac 1 2}\ln(u^2 + \frac 3 4)$, then continue from there.

I don't understand why the integral couldn't become an inverse tangent in this case and why the solution specifically resulted in a logarithm instead when the integral's form perfectly seemed like it'd become the inverse tangent. I understand if I had instead made my substitution $u = x^2 + \frac 3 4$ it would indeed become a natural log, but was what I did truly wrong? Why?

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  • $\begingroup$ If I understand your question: An $x^2+1$ in the denominator of the integrand doesn't guarantee an inverse tangent... Perhaps the clearest example is $$ \int {x^2 + 1 \over x^2 +1 } \,dx = x + C.$$ $\endgroup$ Commented May 27, 2022 at 13:58
  • $\begingroup$ the first one is $\frac{u'}{u}$ this is xhy it integrates in $\ln$. If you do a tan substitution in the first one, the $x$ on the numerator gets in the way (i.e. does not simplifies). $\endgroup$ Commented May 27, 2022 at 13:59
  • $\begingroup$ @peterag I understand that, but the substitution I performed simply made it look to me like it'd be an inverse tangent since the substitution appeared to force it into the inverse tangent form $\endgroup$ Commented May 27, 2022 at 13:59
  • $\begingroup$ @HydroPage then explicitly, what was your substitution? It looks like you took $x^2=u^2$, and figured that made $x\,dx$ into something like $du/2$ ... which doesn't work: with that substitution, $2x \,dx = 2u\,du$. $\endgroup$ Commented May 27, 2022 at 14:04
  • $\begingroup$ @peterag I took $u = x^2$ to cancel the $x$ on top and appear to force the integral into inverse tangent form in terms of $u$. I don't understand why this is incorrect $\endgroup$ Commented May 27, 2022 at 14:05

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Two separate issues. The reason why trig substitution does not work out well is because you have the extra $x$ in the denominator, which makes things more complicated. For example, with typical substitution $x = \tan u$, we have $$ \frac{dx}{x^2+1} = \frac{\sec^2 u \ du}{\tan^2 x + 1} = \frac{\sec^2 u \ du}{\sec^2 u} = du, $$ which makes things nice and clean. But introducing the extra $x$ now has $$ \frac{x dx}{x^2+1} = \tan u \frac{\sec^2 u \ du}{\tan^2 x + 1} = \tan u \frac{\sec^2 u \ du}{\sec^2 u} = \tan u \ du, $$ and you have to integrate the tangent, getting $$ \int \tan u \ du = \int \frac{\sin u}{\cos u} du = - \ln |\cos u| + C $$ which you have to express back in terms of $x$, resulting in a similar rational function.

On the other hand, the second integral is much more pleasant with a direct substitution, since the very $x$ that makes the trig integral unpleasant opens the way to a different more elegant approach which could not be accomplished without it, as is very frequently the case in Mathematics (and art as well).

Consider $u = x^2+1$ then $du = 2x \ dx$ and the integral becomes $$ \int \frac{x\ dx}{x^2+1} = \frac12 \int \frac{du}{u} = \frac12 \ln |u| + C = \frac12 \ln \left|x^2+1 \right| +C $$

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  • $\begingroup$ What I don't understand is why substituting for $x^2$ cancelled out the $x$ on the numerator, but then didn't let me proceed to convert to inverse tangent. Read the last sentence of my question $\endgroup$ Commented May 27, 2022 at 14:03
  • $\begingroup$ @HydroPage you can see from the argument above that in the first case you integrat $\int du = u = \tan^{-1} x$ but in the second one, you end up with $\int \tan u \du$, from the extra $x$ factor in the numerator, that's not so simple to integrate (but possible as the above argument shows -- and you end up in the same place but much lengthier) $\endgroup$ Commented May 27, 2022 at 14:05

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