3 different ways to integrate $ \ln(\sqrt{x})$ and outcomes $3$ different answers.

Question

Method 1

$\int \frac {\ln\sqrt{x}}x$

Let $u = x^{1/2}$ , so $du = \frac 1{2\sqrt{x}}dx$

$2\int \frac {\ln{u}}u$

= $\frac{ln{(u)}^2}3$ = $\frac{ln(\sqrt{x})^2}3$

Method 2

$\int \frac {\ln\sqrt{x}}x$

$\frac12\int\frac{ln{(x)}}x$ = $(ln{(x)})^2-\int\frac{\ln(x)}x$

$\frac32\int\frac{ln{(x)}}x$ = $(ln{(x)})^2$

$\int\frac{ln{(x)}}x$ = $\frac{2(ln{(x)})^2}3$

Method 3

$\int \frac {\ln\sqrt{x}}x$

Let $u = \ln{(\sqrt{x})}$, so $du = \frac1{\sqrt{x}}\times\frac1{2\sqrt{x}}dx$ = $\frac1{2x}dx$

$2\int u du = u^{2}$

= $(\ln{\sqrt{x}})^2$

How's that possible? Different methods give me $3$ different answers? Which one of these is the correct answer?

Answer 1

Apparently none of them, since if you take $u = \ln x$ then $du = \frac{1}{x} dx$, then

$$\frac{1}{2}\int u \ du = \frac{u^2}{4} + C = \color{#f05}{\frac{\ln^2 x}{4}} + C$$

Note: To check your answer compute the derivative of each one.

Answer 2

One has $$\int\frac{\ln\left(\sqrt{x}\right)}{x}dx=\frac{1}{2}\int\frac{\ln\left(x\right)}{x}dx=\frac{1}{2}\int\ln\left(x\right)d\left(\ln\left(x\right)\right)=\frac{1}{4}\left(\ln\left(x\right)\right)^{2}.$$

When you change your variable $u\mapsto f\left(x\right)$, you must be able to express $x\mapsto f^{-1}\left(u\right)$. Your first calculus is wrong because of the factor $\frac{1}{3}$. In your second one, you did and integration by part, but forgot the factor $\frac{1}{2}$. Finally, in the last calculus, you must express $x$ in term of $u$ which may involve some exponential terms maybe...