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I have been doing Integration for quite a while now. And I have improved considerably over the last month. I know u-substitution, IBP and Partial fraction decomposition. Thats it :(

I was just randomly scrolling through the feed a couple of days back and I saw someone substitute something really crazy. It was an entire fraction with not so common terms.

Can anyone give me examples of such awesome integrals, where you have to substitute an entire fraction or an expression? I'd love to give it a try. Also give hints on what to substitute, just in case.

Thanks! Help is appreciated.

William
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  • This might be interesting for you: https://math.stackexchange.com/questions/1075663/trigonometic-substitution-vs-hyperbolic-substitution You can search on Google for relevant exercises to use hyperbolic and trig substitution. – Václav Mordvinov May 11 '18 at 06:41
  • This isn't quite what you're looking for, but I can't resist the opportunity to post it: https://mathoverflow.net/questions/171733/is-it-possible-to-express-int-sqrtx-sqrtx-sqrtx1dx-in-elementary-fun/171856#171856 – Jesse Madnick May 11 '18 at 06:42
  • Also, if you want to increase your integral skills, a challenging book might be good, you can look on this site for reference request, I saw this one a time ago https://math.stackexchange.com/questions/2757186/reference-for-cool-exercises-on-integrals – Václav Mordvinov May 11 '18 at 06:44

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Examples of these "weird fractional substitutions" show up when dealing with integrals involving the $\arctan x$ function. For example, the integral$$I=\int\limits_0^2 dx\,\frac {\arctan x}{x^2+2x+2}$$

Can be solved by using a simple substitution $x\mapsto\frac {2-x}{1+2x}$. The denominator stays the same, and using the identity$$\arctan x+\arctan\left(\frac {2-x}{1+2x}\right)=\arctan 2$$The answer is$$I=\frac 12\arctan 2\arctan\left(\frac 12\right)$$

Or similarly, the integral$$I=\int\limits_0^1 dx\,\frac {\arctan x}{\sqrt{x(1-x^2)}}$$Which can be solved using the substitution $x\mapsto\frac {1-x}{1+x}$ and then using$$\arctan x+\arctan\left(\frac {1-x}{1+x}\right)=\frac {\pi}4$$

The result can be evaluated in terms of the beta function.

Frank W
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