I'm wondering how I can integrate y/(1 + y^2) ... I'm sure there is some trick to doing this but I can't figure it out at the moment.
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I guess it's $\displaystyle \int{\frac{y}{1+y^2}dy}$ isn't it? See it this way:$\displaystyle \int{\frac{y}{1+y^2}dy}=\frac{1}{2}\cdot{\int{\fra c{(1+y^2)'}{1+y^2}dy}}$
More generally: $\displaystyle \int {\frac{{f'(x)}} {{f(x)}}\,dx} = \ln \left| {f(x)} \right|+k\,,\forall f(x) \ne 0$
Originally Posted by PaulRS I guess it's $\displaystyle \int{\frac{y}{1+y^2}dy}$ isn't it? See it this way:$\displaystyle \int{\frac{y}{1+y^2}dy}=\frac{1}{2}\cdot{\int{\fra c{(1+y^2)'}{1+y^2}dy}}$ So, the answer would be 1/2 * ln|1 + y^2| ?? I suppose that works now that I works backwards and derive it. Still, that's a pretty tricky question to ask. =/ Thanks guys!
Yes. -- You can remove the bars 'cause $\displaystyle 1+y^2$ is always positive.
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