Question

Integration!
\[\int \frac{x^2 + 6}{x + 4} dx\]

Answer 1

divide and then integrate term by term...

Answer 2

Ok, let me clean this up a little.
\[\int \frac{x(x + 4) - 4x + 6}{x + 4}\]\[\int \frac{x(x+4) - 4(x + 4) + 16 + 6}{x + 4}\]
That's what you meant, right?
So I end up with
\[\int x - 4 + \frac{22}{x + 4} dx\]

Answer 3

\[\frac{x^2}{2} - 4x + \int \frac{22}{x+4}\]
(I'll do the other part in a sec)

Answer 4

\[\int\limits \frac{ x^2+6 }{ x+4 }\,dx=\int\limits x\,dx - \int\limits 4\,dx +\int\limits \frac{ 22 }{x+4 }\,dx\]