Evaluating $\int_{-1}^{1}x^2\mathrm d(\ln x)$ I am supposed to evaluate the integral $\int_{-1}^{1}x^2\mathrm d(\ln x)$. I solved it traditionally by simplifying the differential to $1/x . \mathrm dx$. This gives the answer as $0$. But as per the solutions, the following answer has been given to be correct.
$$\int_{-1}^{1}x^2\mathrm d(\ln x)=\int_{1/e}^{e}x^2 \frac{1}{x}\mathrm dx=\frac{1}{2}\left(e^2-e^{-2}\right)$$
I do not see why the solution should be this. Besides with a logarithm the limits of integration are $-1$ to $1$, which seems problematic. Could anyone confirm which is correct. Thanks.
 A: Note: In this problem, $\ln x \in [-1,1]$ which implies $x \in [1/e,~e]$ (not $x \in [-1,1]$).
Make a substitution $(t = \ln x)$ to see what is happening:
$$\int_{-1}^{1}x^2\mathrm d(\ln x)=\int_{-1}^{1}x^2 \mathrm dt\overset{t~\to~ x}=\int_{1/e}^{e}x^2 \frac{1}{x}\mathrm dx=\frac{x^2}{2}\bigg|_{1/e}^e=\frac{e^2}{2}-\frac{1}{2e^2}$$
A: @VIVID is right. Integral notation can be misleadingly concise at times, when compared with sums such as $\sum_{i=A}^{i=B}h(i)$. We need to put the equivalent of $i=$ into the limits here. When we write $\int_a^bf(x)dg(x)$,the convention is that it means $\int_{g(x)=a}^{g(x)=b}f(x)dg(x)$, which for strictly increasing $g$ becomes $\int_{x=g^{-1}(a)}^{x=g^{-1}(b)}f(x)g^\prime(x)dx=\int_{g^{-1}(a)}^{g^{-1}(b)}f(x)g^\prime(x)dx$.
Most of the time, you only encounter the case $g(x)=x$. In fairness, some sources will intend limits $a,\,b$ for $x$ instead of $g(x)$. That wouldn't work here, though, because $\ln x$ is undefined in the reals for $x\le0$.
A: we have:
$$I=\int_{-1}^1x^2d(\ln x)$$
and remember that we can write this as:
$$d(\ln x)=\frac{d(\ln x)}{dx}dx=\frac1xdx$$
however, we must also consider that we have:
$$(\ln x)\in[-1,1]\Rightarrow x\in[e^{-1},e]$$

Another way of thinking about this problem would be using substitution, or something which i find a nice visualisation:
$$x=e^{\ln x}\Rightarrow x^2=\left(e^{\ln x}\right)^2$$
so we have:
$$I=\int_{-1}^1\left(e^{\ln x}\right)^2d(\ln x)=\left[\frac{e^{2\ln x}}{2}\right]_{\ln(x)=-1}^1$$

Also, try plotting a graph with $\ln x$ on the x-axis and $x^2$ on the y-axis and see what shape this forms, hope this helps :)
A: I disagree with all the answers and agree with the OP's point-of-view.
It can be argued that the independent variable is $x$ and the integral should/could have expressly been written as
$$\int_{x=-1}^{x=1} x^2 d \ln x = \int_{x=-1}^{x=1} x\,  dx =0.$$
After all, I've seen this notation used as shorthand for integration-by-parts, and the limits referred to the independent variable ($x$) and not to a function of the variable.
The fact that $\ln x$ is not defined for $x=-1$ does not exclude the possibility that the integral is a line integral in the complex plane.  Perhaps a principal value.  Then we have other concerns (which are not insurmountable), branch points and branch cuts ...
