Question

calculus help: integrate (x+4)/x^2+2x+5

Answer 1

I got stuck at (x+4)/((x+1)^2+4)

Answer 2

See how you have an (x+4) in the numerator?
I think you want to try and get an (x+4)^2 in the denominator.

It will make a substitution easier to accomplish.
Looks like we'll apply a trig sub after you get that sorted out.

Answer 3

I was also thinking of doing -4 + 4 in the numerator and breaking it into two integrals.

Answer 4

oops.

Answer 5

nvm x_x

Answer 6

yah looks like that first one was gonna give us a lil trouble :c

Answer 7

I just noticed, haha. Yeah, trig-sub I like then.

Answer 8

Oh we can't do an (x+4)^2 in the denominator can we? :P derp! my bad.

Answer 9

We can do trig-sub even though we don't have a square root, right?

Answer 10

\[\large \int\limits \frac{x+4}{(x+1)^2+4}dx\]

Yah we don't need a square root :)
Hopefully this will work out.. hmm
Making the substitution:\[\large x+1=2\tan \theta\]

Answer 11

\[\large \int\limits \frac{(x+1)+3}{(x+1)^2+4}dx\qquad=\qquad \int\limits\frac{(2\tan \theta)+3}{4\tan^2\theta+4}dx\]Factoring out a 4 in the denominator,\[\large \frac{1}{4}\int\limits \frac{2\tan \theta+3}{\tan^2\theta+1}dx \qquad=\qquad \frac{1}{4}\int\limits\frac{2\tan \theta+3}{\sec^2\theta}dx\]Then we still need to deal with the dx.

\[\large x+1=2\tan \theta \qquad\to\qquad dx=2\sec^2\theta\;d \theta\]

Giving us,\[\large \frac{1}{4}\int\limits\limits\frac{2\tan \theta+3}{\cancel{\sec^2\theta}}(2\cancel{\sec^2\theta}\;d \theta)\]

Answer 12

Ok I'll simmer down, I don't wanna do all it and take the fun away from ya :O
Do you see where I'm going with this though?
Stuck on any step in there?

Answer 13

Hmm...Lol, didn't even need to do what I'd have done. Yay for more than one way to do the trig sub.

Answer 14

\[\int\frac{x+4}{x^2+2x+5}~dx=\frac{1}{2}\left(\int\frac{2x+2}{x^2+2x+5}~dx\right)+\int\frac{3}{(x+1)^2+4}~dx\]
Not much different from what the others suggested, since it involves an algebraic and trig substitution, but the trig sub would be easier on the eyes.