Integral of (x+3)/(x^2+4) I'm trying to do it by parts: u = x+3 du = 1 dv = 1/(x^2+4) v = 1/2 arctan (x/2) but, I'm not sure this is the right way. It gets a bit messy and I assume I'm doing something wrong.
Follow Math Help Forum on Facebook and Google+
We may use the fact that $\displaystyle \frac{x+3}{x^2+4}=\frac{x}{x^2+4}+\frac{3}{x^2+4}.$
View Tag Cloud