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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).

user84413
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2 Answers2

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$$\begin{align}\int\frac{1}{(t^2+25)^2}&=\frac{1}{25}\int\frac{t^2+25-t^2}{(t^2+25)^2}\\&=\frac{1}{25}\int\frac{1}{t^2+25}+\frac{1}{2\cdot25}\int t\cdot\frac{ 2t}{(t^2+25)^2}\end{align}$$

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 $k$.

Pp..
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  • Very cool. An alternative method to trig...Didn't know that. – imranfat Jan 30 '15 at 18:10
  • @imranfat There was a time before James Stewart built his 29 million dollar house out of selling books of ignorance, in which this wouldn't impress you. This was the normal approach to integrating this case of simple fractions. Trigonometric substitutions may have impressed you. As they are only useful to get faster answers to a handful of integrals. They were taught as a curiosity or not taught. – Pp.. Jan 30 '15 at 18:30
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$$ \begin{aligned} \int \frac{1}{\left(t^{2}+25\right)^{2}} d t &=-\frac{1}{2} \int \frac{1}{t} d\left(\frac{1}{t^{2}+25}\right) \\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{2}\left(\frac{1}{t^{2}\left(t^{2}+25\right)} d t\right.\\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50} \int\left(\frac{1}{t^{2}}-\frac{1}{t^{2}+25}\right) d t \\ &=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50}\left[-\frac{1}{t}-\frac{1}{5} \tan ^{-1}\left(\frac{t}{5}\right)\right]+C \\ &=\frac{1}{250}\left[\frac{5}{t^{2}+25}+\tan ^{-1}\left(\frac{t}{5}\right)\right]+C \end{aligned} $$

Lai
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