# Integration by inspection - how?

The textbook says that

$$\int\frac{-mu}{\sqrt{1-\left(\frac{u}{c}\right)^{2}}}du$$
can be solved by inspection to give $$mc^{2}\sqrt{1-\left(\frac{u}{c}\right)^{2}}$$

In simple steps, could anyone please explain how this is done.

Thank you

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The integrand has the form $\frac{f'}{\sqrt f}$ (up to a multiplicative constant), so it's the derivative of $\sqrt f$. –  Davide Giraudo Sep 28 '11 at 11:13
At least by inspection one can see that the answer is right! To proceed in the other direction, one notices that the derivative of $1-(u/c)^2$ is (basically) sitting upstairs, so the answer will be, more or less, $\sqrt{1-(u/c)^2}$. Then I would mentally differentiate to decide on the constant. –  André Nicolas Sep 28 '11 at 14:21
Well, integration by inspection proceeds via the following simple steps: (1) Inspect. (2) Integrate. :-) –  Hans Lundmark Sep 28 '11 at 17:51

Note that the integrand is proportional to $f'(g(u))g'(u)$ with $f(x)=\sqrt x$ and $g(u)=1-\left(\frac uc\right)^2$. Generally, if an integrand is a product of some expression and the derivative of some part of that expression, try substituting for that part.