integral of (u+4)/((u^2)+4) du

How would I solve for this?

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- Jun 6th 2008, 09:40 AMkhuezyintegration partial fractions
integral of (u+4)/((u^2)+4) du

How would I solve for this? - Jun 6th 2008, 09:43 AMMathstud28
THere is no need for partial fractions here

$\displaystyle \int\frac{u+4}{u^2+4}du=\int\frac{u}{u^2+4}du+4\in t\frac{du}{u^2+4}=\frac{1}{2}\ln|u^2+4|+2\arctan\b igg(\frac{u}{2}\bigg)+C$

Also, conider the fact that if you wanted to do this by partial fractions the denominators factorization would be imaginary

And even if it is too hard to learn LaTeX a useful command is alt 0178 gives ² - Jun 6th 2008, 09:57 AMKrizalid
Since $\displaystyle u^2+4$ is positive for all $\displaystyle u,$ we can remove the absolute value bars.

- Jun 6th 2008, 09:58 AMMathstud28