Question

Session75 Recitation Part C

Answer 1

I keep coming up with \[\int\limits_{}^{}\frac{(2x+2)dx}{(4x^2+1)^2} = \frac{x^2+2x}{1+4x^2} + \frac{arctan(2x)}{2} + C\]

but online integration calculator returns \[\frac{arctan(2x)}{2} + \frac{4x-1}{16x^2+4} +C \]

So I have the arctan(2x)/2 in common, but my other part doesn't match.

I first substitute u = 2x, du = 2dx...
\[\frac{(u+2)du}{2(u^2+1)^2}\]

And then use trig substitution to set... \[u = \tan(\theta), du = \sec^2(\theta)d\theta\]

Answer 2

Bah, simple algebra mistakes. CLOSED.