Vicious circle with $\int_1^2{\ln x \over x}{dx}$

Question

Calculate $$E(Ω)=\int_1^2{\ln x \over x}{dx}$$ $E$ derives from the greek word "εμβαδόν" which in English means area. $Ω$ is the specific "area" that is being calculated by the integral.

Personal work:

$$E(Ω)=\int_1^2{\ln x \over x}{dx}= \int_1^2{\ln x {1\over x}}{dx}=\int_1^2{\ln x (|\ln x|)'}{dx}=[\ln x|\ln x|]_1^2-\int_1^2(\ln x)'|\ln x|dx=\cdots=\ln4-\int_1^2 {1\over x}|\ln x|\,dx$$

I've been taught in school that when I have an integral form like this: $P(x)*\ln(kx)$ I always rewrite the $P(x)$ as a derivative. My problem is that it is a vicious circle and I always get $$\int_1^2{\ln x}{(|\ln x|)'}.$$ What am I doing wrong?

Answer 1

Put $u=\ln x$ hence $du=\frac {dx}{x}$

The integral changes to $$\int_0^{\ln2} udu=\frac {u^2}{2}=\frac {(\ln 2)^2}{2}$$

Answer 2

Hint: Substitute $$t=\ln x$$ then you will get $$\mathrm dt=\frac{1}{x}\,\mathrm dx$$

Answer 3

You have just concluded that if $$I = \int_{x=1}^2 \frac{\log x}{x} \, dx,$$ then $$I = (\log 2)^2 - I.$$ Consequently, $$2I = (\log 2)^2$$ and $$I = \frac{(\log 2)^2}{2}.$$

Note that your original result is not correct; $(\log 2)^2 \ne \log 4$.

As others have commented, an approach via direct substitution and not integration by parts is simpler and achieves the same result.