How to solve integral of rational function with logarithm

Question

I have no idea how to solve this integral: $$ \int\frac{dx}{x\sqrt{1−\ln^2(x)}} $$ What should I substitute?

Answer 1

You should substitute $u=\ln(x)$ because $du=1/x \, dx$ appears elsewhere in the integrand. If we write the integral as $$ \int \frac{1}{\sqrt{1-\ln(x)^2}} \cdot \frac{1}{x} \, dx \, $$ then it is clear that when we make the substitution $u=\ln(x)$ it simplifies nicely $$ \int \frac{1}{\sqrt{1-u^2}} \, du \, . $$

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Integration by substitution comes from reversing the chain rule. Recall that if $y=f(g(x))$, then $$ \frac{dy}{dx}=f'(g(x))g'(x) \, . $$ Thus, $$ \int f'(g(x))g'(x) \, dx = f(g(x))+C \, . $$ On the other hand, if we make the substitutions $u=g(x)$ and $du = g'(x) \, dx$, then the integral becomes $$ \int f'(u) \, du=f(u)+C=f(g(x))+C \, . $$ This shows that when we make a substitution of the form $u=g(x)$, ideally $g'(x)$ should appear in the integrand. It doesn't have to, though. Consider the integral $$ I = \int \frac{1+e^x}{1-e^x} \, dx \, . $$ At first sight, it doesn't seem like the substitution $u=e^x$ will work, or is even mathematically correct. However, if we rewrite $I$ as $$ I = \int \frac{1+e^x}{1-e^x} \cdot \frac{1}{e^x} \cdot e^x \, dx $$ then $I$ can be expressed in the form $\int f'(g(x))g'(x) \, dx$! So every substitution ultimately boils down to the reverse chain rule.

There are even more algebraic tricks you can use to evaluate integrals; if you are interested in hearing them, just ask.

Answer 2

$$u=\ln x \implies du=\frac{dx}{x}$$

$$∫\dfrac{dx}{x\sqrt{(1−ln^2(x))}}=∫\dfrac{du}{\sqrt{1−u^2}}=\arcsin u +c= \arcsin (\ln x) +c$$