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?
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 $$ \int \frac{x\ dx}{x^2+1} = \frac12 \int \frac{du}{u} = \frac12 \ln |u| + C = \frac12 \ln \left|x^2+1 \right| +C $$