You don't need to use integration by parts. Consider
$$z=x^2+1,$$
then,
$$dz=2xdx.$$
Thus,
$$\int 2x\cos(x^2+1)dx=\int \cos(z)dz=\sin(z)=\sin(x^2+1).$$
Riemann integrals $\int g(x)dx$ are limits of sums with ever narrower area strips approximated as with-$dx$ rectangles. This affords a generalization: $\int_a^bf(x,\,dx)$, if it exists, is $\lim_{n\to0}\sum_{i=0}^{n-1}f(a+i/n,\,(b-a)/n)$.
(We want the same limit to be obtained for arbitrary ways of splitting $[a,\,b]$ into $n$ strips, not just those where their widths are equal; the definition of Riemann integrals already encounters this subtlety, but we'll overlook it herein because it's more instructive for now to illustrate the limit in the special case than to verify the general case obtains it.)
In your example,$$\int_a^b(x^{dx}-1):=\lim_{n\to0}\sum_{i=0}^{n-1}((a+i/n)^{(b-a)/n}-1)=\lim_{n\to0}\frac{b-a}{n}\sum_{i=0}^{n-1}\ln(a+i/n)=:\int_a^b\ln xdx,$$where $u:=v,\,v=:u$ both mean "$u$ is defined as $v$". Or indefinitely, $\int x^{dx}-1=\int\ln xdx$ (which makes sense in light of $x^h-1=e^{h\ln x}-1\sim h\ln x$ for small $h$).
You can use this kind of reasoning to show that, if $f(x,\,h)\sim g(x)h$ for small $h$, $\int f(x,\,dx)=\int g(x)dx$. You may want to consider other examples, such as $\lim_{h\to0}f(x,\,h)=0$ or $f(x,\,h)\sim g(x)h^2$.
Best Answer
Comparing Taylor series with $df=f^\prime dx$ gives $dx^2=0$. Note that$$g(x)^{dx}-1=\exp(\ln g(x)\cdot dx)-1=\ln g(x)\cdot dx+O(dx^2)=\ln g(x)\cdot dx,$$so your first example is $\int\ln xdx=x\ln x-x+C$.