Question
Evaluate $$\int_{2}^{4} \frac {\mathrm{d}x} {x \ln x}\ .$$
My working
Let $u = \frac {1} {\ln x}$ and $v' = \frac {1} {x}$
$\implies u' = -\frac {1} {x (\ln x)^2}$ and $v = \ln x$
$\therefore \int_{2}^{4} \frac {dx} {x \ln x} = 1 + \int_{2}^{4} \frac {dx} {x \ln x}$
$\implies 0 = 1$ (say what?)
Answer
$\int_{2}^{4} \frac {dx} {x \ln x} = \ln 2$
When I use the substitution $u = \ln x$ and proceed, I do arrive at the answer, but that is trivial so I am not here to discuss that. What I am here to discuss, however, is my working when I use integration by parts. I seem to have gone wrong somewhere, which I find very intriguing. I assume I must have been careless, but I have been doing calculus all day, so perhaps my mind is fatigued. Even worse, is there some inherent misunderstanding in my concept of integration by parts? I will be very grateful if anyone can point out where I have gone wrong 🙂
Edit
Following the answers given, it seems I did have a conceptual misunderstanding about integration by parts and it turns out that integration by parts cannot be used to solve this particular integral! Today, I have also found out that integration by parts cannot solve all integrals!
Best Answer
Integration by parts states $$\int_2^4 uv' dx = uv\big|_2^4 - \int_2^4 u'v dx$$
We have $uv=1$. Hence $uv\big|_2^4 = 0$.
The equation becomes $$\int_{2}^{4} \frac {dx} {x \ln x} = 0 + \int_{2}^{4} \frac {dx} {x \ln x}$$
and no conclusion can be drawn.