Evaluating the Integral: $\int \left(\frac{\ln(\sqrt{x})}{x} \right) \, dx$

Question

I have to evaluate the integral $\int \left(\frac{\ln(\sqrt{x})}{x}\right)\,dx$.

My calculator gives the answer $\frac{(\ln(x))^2}{4}+C$

When I began integrating I used integration by parts:

So I began by first bring out the half from the integral

$$\int\left(\frac{\ln(x^{\frac{1}{2}})}{x}\right) \,dx$$ $$\frac{1}{2}\int\left(\frac{\ln(x)}{x}\right)\,dx$$ Then I set my $u$ and $dv$ for integration: $u=\frac{1}{x}$, and $du=-\frac{1}{x^2} \, dx$, $dv=\ln(x) \, dx$, and $v=x\ln(x)-x$

$$I=\frac{1}{2} \left[ \frac{1}{x}(x\ln(x)-x)-\int\left(-\frac{1}{x^2}\right)(x\ln(x)-x) \right]$$

$$I=\frac{1}{2} \left[(\ln(x)-1)-\int-\left(\frac{\ln(x)}{x}-\frac{1}{x}\right) \, dx \right]$$

I automatically know this has to be wrong because it does not work, but does work?

Answer 1

$$\frac{1}{2} \int \frac{\ln(x)}{x} dx$$ $u=\ln(x)$ and $du=\frac{1}{x}$ so then it is now $$\frac{1}{2}\int u du= \frac{u^2}{2} \frac{1}{2}+C=\frac{u^2}{4}+C$$ subtitute $u$ back and the answer is $\frac{(\ln(x))^2}{4}+C$

Answer 2

With $u=\ln x$, then $du=\dfrac{1}{x} \, dx$, so $\displaystyle\int\dfrac{\ln x} x \, dx=\int u\,du=\dfrac{1}{2} u^2+C$.