Question

Partial fraction decomposition:

int [u/(u^2 - u +1)] du

Answer 1

I don't think that's the way to go with this problem. If it's not required, I suggest the following:
\[\int\frac{u}{u^2-u+1}~du=\int\frac{u-1}{u^2-u+1}~du+\int\frac{1}{u^2-u+1}~du\]
For the first integral, substitute \(t=u^2-u+1\) so that \(\dfrac{1}{2}dt=u~du\), and for the second, complete the square in the denominator:
\[\frac{1}{2}\int\frac{dt}{t}+\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}~du\]
The proper trig sub would be \(u-\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\tan p\) so that \(du=\dfrac{\sqrt{3}}{2}\sec^2p~dp\):
\[\frac{1}{2}\int\frac{dt}{t}+\int\frac{1}{\left(\frac{\sqrt{3}}{2}\tan p\right)^2+\frac{3}{4}}~\left(\frac{\sqrt3}{2}\sec^2p~dp\right)\]
\[\frac{1}{2}\int\frac{dt}{t}+\frac{4}{3}\cdot\frac{\sqrt3}{2}\int\frac{\sec^2p}{\tan^2 p+1}~dp\]

Answer 2

hmm , for the first suggestion - if you subtract one shouldnt you have added one , so that the overall expression remains unchanged

Answer 3

I appreciate the help nonetheless , but It wasnt the correct method of approach

Answer 4

I did add one. That's where the second integral comes from.