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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!

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    $\begingroup$ $$\int\dfrac{dx}{\sqrt x}=?$$ $\endgroup$ Commented Sep 30, 2019 at 12:11
  • $\begingroup$ A comment is highly appreciated upon putting a downvote. Otherwise, there is no way for me to know what is wrong with the question and maintain high-quality posts. $\endgroup$
    – roman
    Commented Sep 30, 2019 at 12:28

1 Answer 1

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

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  • $\begingroup$ Oh, thank you very much! shame on me $\endgroup$
    – roman
    Commented Sep 30, 2019 at 12:15

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