Our calculus book covers partial fractions but not trig substitution, so I would like to find out the most elementary way to evaluate
$$\displaystyle\int\frac{1}{(t^2+25)^2}\;dt$$
without using trig substitution (or partial fractions over the complex numbers).
$$\begin{align}\int\frac{1}{(t^2+25)^2}&=\frac{1}{25}\int\frac{t^2+25-t^2}{(t^2+25)^2}\\&=\frac{1}{25}\int\frac{1}{t^2+25}+\frac{1}{2\cdot25}\int t\cdot\frac{ 2t}{(t^2+25)^2}\end{align}$$
The $t$ disappears by differentiation and the $\frac{2t}{(t^2+25)^2}$ integrates to $\frac{1}{t^2+25}$. So, integration by parts with that last integral.
The same idea allows you to integrate the simple fractions of the form $$\frac{A}{(x^2+px+q)^k}$$ by reducing the $k$.
$$ \begin{aligned} \int \frac{1}{\left(t^{2}+25\right)^{2}} d t &=-\frac{1}{2} \int \frac{1}{t} d\left(\frac{1}{t^{2}+25}\right) \\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{2}\left(\frac{1}{t^{2}\left(t^{2}+25\right)} d t\right.\\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50} \int\left(\frac{1}{t^{2}}-\frac{1}{t^{2}+25}\right) d t \\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50}\left[-\frac{1}{t}-\frac{1}{5} \tan ^{-1}\left(\frac{t}{5}\right)\right]+C \\ &=\frac{1}{250}\left[\frac{5}{t^{2}+25}+\tan ^{-1}\left(\frac{t}{5}\right)\right]+C \end{aligned} $$
