How do you integrate #f(x)=x/(x^2+1)# using the quotient rule?

1 Answer
Jan 15, 2017

You throw the quotient rule away and write down #1/2 ln(x^2+1)#

Explanation:

The quotient rule is from differentiation, not integration.
You can guess that the answer is something like #ln(...)# because the differentiation #ln(...)# gives something like #1/(...)#. You then spot that the top is (give or take a constant factor) what you get by differentiating the bottom, #(x^2+1)# yielding #2x#. So try differentiating #ln(x^2+1)# using the chain rule and compare what you get with the question. In essence, you have to spot the `reverse chain rule'.

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Notice how for large values of #x#, both positive and negative, the answer gets close to #log|x|#. This is because the #1# becomes negligible compared to the #x^2#, so the answer gets close to #(1/2)ln(x^2)#, which is #ln |x|#.

In the graph above, the green line is the exact integral of the red line, and the blue line is the approximation achieved by ignoring the #1#, for suitably chosen values of the constant of integration. If you imagine these graphs being repeated with successively smaller values instead of #1# (say #0.1#, #0.01#, ... ) then the red graph sort of approaches the curve #y=1/x# (which has a vertical asymptote #x=0#), but "panics" at the thought of infinity and takes a short cut through the origin to the corresponding point on the other side!