Can integration by parts be used to integrate functions such as $f(x) = x^3$ or $f(x) = \frac 1x$?
I know that there are simple formulas that can be used to integrate functions of those forms already, but could we use integration by parts?
I tried integrating $f(x) = x^3$ with integration by parts, where $u = x^3$ and $dv = dx$. But it ended up being an infinite loop of integrals, although, I may have done something wrong.
The reason I bring this up is because I saw an example where the integral of $\ln(x)$ was solved to get $x\ln(x) - x$ by using integration by parts where $u = \ln(x)$ and $dv = dx$. So shouldn't we be able to apply this technique to other integrals where $dv = dx$?
In every case,
$$\int f(x)\,dx=xf(x)-\int xf'(x)\,dx$$
and this is useful when $xf'(x)$ is easier to integrate than $f(x)$.
In the case of powers of $x$, you indeed seem to circle in rounds because the integrand keeps being the same, but you work around the circularity with
$$\int x^r\,dx=x\,x^r-r\int x\,x^{r-1}\,dx$$ and $$\int x^r\,dx+r\int x^r\,dx=(1+r)\int x^r\,dx=x^{r+1}.$$
Notice that this doesn't work for $r=-1$, which is precisely the case where the logarithm enters into play. (In some sense $x^0\equiv \log x$.)