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I was calculating the integral $\int \frac{x}{\sqrt{1-x^2}} \space dx$ and noticed by accident that $\frac{\text{d}}{\text{d}x}\sqrt{1-x^2}=-\frac{x}{\sqrt{1-x^2}}$. This allowed me to calculate the integral, but the only reason I did the differentiation was to try to somehow fit the integral into $\int \frac{f'(x)}{f(x)} \space dx$. Are there some hints I could look for to notice these, or just test/know them?

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    I guess this one of those cases where the old, unsatisfying "you'll develop an intuition for it over time" answer makes an appearance.2017-01-06

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It doesn't fit in the $f'/f$ scheme but in the chain rule scheme: $$\int (f\circ g)\cdot g'=F\circ g.$$

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First you have to realise that the derivative of sqrtx is 1/2 sqrt(1/x)

and by the chain rule that the derivative of x^2 is 2x

Integrals are, by their nature, difficult