3
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Still reading. First solution seems good. $\endgroup$ Feb 22 '19 at 6:20
  • 1
    $\begingroup$ Note the final integral of $\tan(x)$ is $\ln(\sec(x))$ and $\sec(x)=(\tan^2(x)+1)^{1/2}$...that square root is what is missing $\endgroup$
    – user35508
    Feb 22 '19 at 6:27
  • 1
    $\begingroup$ It's all good until you convert back to x on the last couple lines. You should indeed get that $-\log(\cos(\arctan(\tfrac{x+1}{2}))) = \tfrac{1}{2} \log(x^2+2x+5)$ $\endgroup$ Feb 22 '19 at 6:27
  • 1
    $\begingroup$ $\cos \theta = \frac {2} {\sqrt {u^2+4}}.$ Right? $\endgroup$
    – little o
    Feb 22 '19 at 6:28
3
$\begingroup$

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.

  1. $\ln \left|\sqrt{u^2+4}\right|=1/2\cdot\ln\left|u^2+4\right|$
  2. $1/2\cdot\ln\left|(u^2+4)/\sqrt{2}\right|=1/2\cdot\ln\left|u^2+4\right|-1/2\cdot\ln\sqrt{2}$ which gets absorbed in the constant $C$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.