Final review question

I am having a little trouble with two of the u-substitution problems on the review sheet.

Is this the correct way to solve problem 1(d)?
1 (d)
\int\frac{1}{xln(x)}dx
\int\frac{1}{xln(u)}dx
u=x
\frac{du}{dx}=1
du=dx
\int\frac{1}{xln(u)}du
ln(ln(x))+C

The next one I am unsure about is problem 2(b). Is this the correct way to do this one?

\int_{-1}^1 xe^{sin(x^2)}dx
\int_{-1}^1 xe^{sin(u)}dx
u=x^2
\frac{du}{dx}=2x
\frac{1}{2}du=xdx
\frac{1}{2}\int e^{sin(u)}du
u=(-1)^2 u=(1)^2
\int_{1}^1 \frac{1}{2}e^{sin(u)}
\frac{1}{2}e^{sin(1)}- \frac{1}{2}e^{sin(1)}=0

Here are a couple of hints or suggestions.

For (1d): The substitution u=x is really never useful; it only succeeds in changing the name of the variable of integration. In this particular case, you have

so

Note that you’ve got to get rid of both of the x s in the original integrand.

I think that a better start would be to set u=\ln(x).

In 2b: You’re petty close, though you’ve seem to have dropped the du near the end; your next to last line should be

Your final answer of zero is actually correct, but I don’t think you can find an antiderivative for this problem - at least WolframAlpha can’t. Thus, you’ll need to argue via some other means that value is zero.