What is the wrong step in the integral $ \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx $? 
Evaluate the following integral:
  $$
I= \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx
$$

I've started with a substitution: $t = \ln x$, then:
$$
dt = {dx \over x} \iff dx = xdt\\
I = \int {tdt\over \sqrt{1 - 4t - t^2}}
$$
Completing the square in the denominator I got:
$$
1-4t-t^2 = -(t^2 + 4t - 1 +5-5) = -(t+2)^2 + 5 = 5-(t+2)^2
$$
Then the integral becomes:
$$
\int \frac{tdt}{\sqrt{5-(t+2)^2}}
$$
Substitute $t+2 = s$, then $dt = ds$, and $t = s-2$:
$$
\int \frac{(s-2)ds}{\sqrt{5 - s^2}} = \int \frac{sds}{\sqrt{5 - s^2}} - \int\frac{2ds}{\sqrt{5 - s^2}} \tag1
$$
Then:
$$
I_1 = \int \frac{sds}{\sqrt{5 - s^2}}
$$
Substitute $p = s^2$, $dp = 2sds$, and $ds = {dp \over 2s}$:
$$
I_1 = \int 
\frac{dp}{2\sqrt{5-p}} = {1\over 2}\arcsin{\sqrt{p}\over \sqrt5}+C = \\
{1\over 2}\arcsin{\ln x + 2\over \sqrt5}+C
$$
Going back to $(1)$:
$$
I_2 = \int\frac{2ds}{\sqrt{5 - s^2}} = 2\arcsin{s\over \sqrt5}+C= \\
2\arcsin{\ln x + 2\over \sqrt5}+C
$$
Which means:
$$
I = I_1 - I_2 = \boxed{{1\over 2}\arcsin{\ln x + 2\over \sqrt5} - 2\arcsin{\ln x + 2\over \sqrt5}+C}
$$
And that is not correct since the answer suggests:
$$
I = -\sqrt{1-4\ln x - \ln^2 x} - 2\arcsin{\ln x + 2\over \sqrt5}+C
$$
I've been trying to spot the error for a while without any success, where did it go wrong? Obviously my answer is wrong. By the way, I'm supposed to use substitution to solve the integral. Thank you in advance!

Your computation of $I_1$ is wrong. $\int \frac 1 {\sqrt {5-p}} dp$ is $-2\sqrt {5-p}+C$
