Integrating $\ln x$ by parts I am asked to integrate by parts $\int \ln(x) dx$. But I'm at a loss isn't there supposed to be two functions in the integral for you to be able to integrate by parts? 
 A: $$\int \ln x \, dx= \int 1\cdot \ln x\, dx= x\ln x- \int x\cdot \frac{1}{x}\, dx=x\ln x-\int \,dx = x\ln x-x+C,$$
where $C$ is a constant.
A: $$
\int \ln x\,dx = \underbrace{\int u\,dx = ux - \int x\,du}_{\text{integration by parts}} = x\ln x - \int x\,\frac{dx}{x}.
$$
Now cancel the $x$ from the numerator and denominator and go from there.
(I've seen probably at least a couple of dozen students fail to figure out that a cancelation can be done there.  They wonder about such things as whether one should integrate the $x$ and the $dx/x$ separately and then multiply.)
The arctangent function is done the same way:
$$
\begin{align}
\int\arctan x\,dx & = \underbrace{\int u\,dx = ux - \int x\,du}_{\text{integration by parts}} = x\arctan x - \int x\, \frac{dx}{1+x^2} \\[8pt]
& = x\arctan x - \int\frac{1}{1+x^2} \cdot\frac12\cdot \Big(2x\,dx\Big) \\[8pt]
& = x\arctan x - \frac12\int\frac1w\,dw \\[8pt]
& = x\arctan x-\frac12\ln |w| + C \\[8pt]
& = x\arctan x - \frac12\ln(1+x^2)+C.
\end{align}
$$
A: Just a good point: Wherever you have $$\int p(x)\ln(x)dx$$ in which $p(x)$ is an integrable function; you can use the integration by parts as follows: $$u=\ln(x),~~dv=p(x)dx$$ such that $$\int udv=uv-\int vdu$$ 
A: Hint: Write $\log(x)$ as $1 \cdot \log(x)$ and use integration by parts.
