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

Jan 15, 2017

You throw the quotient rule away and write down $\frac{1}{2} \ln \left({x}^{2} + 1\right)$

#### Explanation:

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

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 $\left(\frac{1}{2}\right) \ln \left({x}^{2}\right)$, 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 $y = \frac{1}{x}$ (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!