When I complete the square in the denominator and solve using u-sub, I can get the right answer:
$$\int\frac{x+4}{x^2+2x+5}dx$$ $$\int\frac{x+4}{x^2+2x+1+4}dx$$ $$\int\frac{(x+1)+4}{(x+1)^2+4}dx$$ $$u=x+1$$ $$\int\frac{u+3}{u^2+4}du$$ $$I_1=\int\frac{u}{u^2+4}du+I_2=\int\frac{3}{u^2+4}du$$ $$I_1=(w=u^2+4, dw=2udu)$$ $$I_1=\frac{1}{2}\int\frac{dw}{w}$$ $w = u^2+4,$
$w = (x+1)^2+4,$ so: $$I_1=\frac{1}{2}\ln|x^2+2x+5|$$ $$I_2=3\int\frac{1}{u^2+4}du$$ $u = 2\tan\theta$
$du = 2\sec^2\theta d\theta$ $$I_2=3\int\frac{2\sec^2\theta}{(2\tan\theta)^2+4}d\theta$$ $$I_2=3\int\frac{2\sec^2\theta}{4\tan^2\theta+4}d\theta$$ $$I_2=3\int\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$$ $$I_2=\frac{3}{2}\int d\theta$$ $$I_2=\frac{3\theta}{2}$$
At this point, I have to turn the integral back into terms of x, so I made the right triangle like normal:
Now solving the integrals:
$$\frac{1}{2}\ln|x^2+2x+5|+\frac{3\theta}{2}$$ $$\frac{1}{2}\ln|x^2+2x+5|+\frac{3\arctan(\frac{x+1}{2})}{2}+C$$
This obviously is the correct answer, however, when I try to solve this with trig-sub (which is the first thing that came to my mind when I looked at the problem, hence my frustration) I am getting a similar, albeit incorrect answer:
$$I_1=\int\frac{u}{u^2+4}du+I_2=\int\frac{3}{u^2+4}du$$ $u = 2\tan\theta$
$du = 2\sec^2\theta d\theta$
$$I_1=\int\frac{2\tan\theta}{4\tan^2\theta+4}\cdot\frac{2\sec^2\theta}{1}d\theta$$ $$I_1=\int\frac{2\tan\theta}{4\sec^2\theta}\cdot\frac{2\sec^2\theta}{1}d\theta$$ $$I_1=\int\frac{2\tan\theta}{2}d\theta$$ $$I_1=\int \tan\theta$$ $$I_2=3\int\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$$ $$I_2=3\int\frac{1}{2}d\theta$$ $$I_2=\frac{3\theta}{2}$$ $$\int \tan\theta d\theta+\frac{3\theta}{2}$$ $$-\ln|\cos\theta|+\frac{3\theta}{2}$$ Now I put it back into terms of x like I did when solving it using u-sub: $$-\ln\left|\frac{2}{u^2+4}\right|+\frac{3}{2}\arctan\frac{x+1}{2}$$ Since $u=x+1$: $$-\ln\left|\frac{2}{(x+1)^2+4}\right|+\frac{3}{2}\arctan(\frac{x+1}{2})+C$$
But this is obviously wrong, since it looks like the $ln$ should have a $\frac{1}{2}$ in front of it, so something must be wrong with my trig-sub on $I_1$? I know it's a lot to read but I just wanted to put it step by step to see if there's some dumb algebraic mistake I made. If anyone can help, thanks a ton in advance.
You plugged in the wrong expression for $\cos \theta$, it should be $2/\sqrt{u^2+4}$. Notice that the Pythagorean theorem tells you that the length of the hypotenuse should be $\sqrt{x^2+2x+5}$.
Note the minus sign outside of the logarithm. $$-\ln\left|\dfrac{2}{\sqrt{u^2+4}} \right|=\ln\left|\dfrac{\sqrt{u^2+4}}{2}\right|$$
Since $\sqrt{f(x)}$ can be written as $(f(x))^{1/2}$. Therefore by the logarithm properties.