# 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?

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Hint: Write $\log(x)$ as $1 \cdot \log(x)$ and use integration by parts.

$$\int \ln x \, dx= \int 1\cdot \ln x\, dx= x\ln x- \int x\cdot \frac{1}{x}\, dx=x\ln x-\int \,dx = x\ln x-x+C,$$
where $C$ is a constant.

$$\int \ln x\,dx = \underbrace{\int u\,dx = ux – \int x\,du}_{\text{integration by parts}} = x\ln x – \int x\,\frac{dx}{x}.$$
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}

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 $$\int udv=uv-\int vdu$$