# 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!

• $$\int\dfrac{dx}{\sqrt x}=?$$ – lab bhattacharjee Sep 30 '19 at 12:11
• 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. – roman Sep 30 '19 at 12:28

## 1 Answer

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

• Oh, thank you very much! shame on me – roman Sep 30 '19 at 12:15