Question

I'm uploading a problem that is all worked out. (It's in the comments as a .png attachment)

Could someone explain the step to me that is boxed in blue? Thank you!

Answer 1

it is algebra

Answer 2

but the point is this
if you change the numerator of
\[\frac{x+6}{x^2+3x+9}\] in to
\[\frac{x+\frac{3}{2}}{x^2+3x+9}\] then \[\int \frac{x+\frac{3}{2}}{x^2+3x+9} dx\] is a real easy \(u\) - sub

Answer 3

\[u=x^2+3x+9,du=(2x+3)dx,\frac{1}{2}du=(x+\frac{3}{2})dx\] and the integral is not simple
\[\frac{1}{2}\int \frac{du}{u}=\ln(u)\]

Answer 4

so the gimmick was to break it in to two pieces, one to make a simple u - sub the the rest is
\[\int\frac{ \frac{9}{2}}{x^2+3x+9}dx\] since
\[6-\frac{3}{2}=\frac{9}{2}\]

Answer 5

Wow. Thank you!