<s>[INST] 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!
 [/INST] Your computation of $I_1$ is wrong. $\int \frac 1 {\sqrt {5-p}} dp$ is $-2\sqrt {5-p}+C$
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