integral of (u+4)/((u^2)+4) du
How would I solve for this?
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
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