Compute the following integral $$\displaystyle\int\ln(x) \ dx$$
Development:
We have to choose a function that is $$u(x)$$ and other one $$v(x)$$, so that $$\ln(x)$$ is expressed like $$\ln(x) =u (x) \cdot v' (x)$$.
We choose in this case: $$$u=\ln(x) \ \ ; \ \ dv=1\cdot dx$$$
And we have $$$du=\dfrac{1}{x} \ \ ; \ \ v=\displaystyle\int 1\cdot \ dx=x$$$
We can now apply the integration by parts formula, and we have:
$$$\int\ln(x) \ dx=\int\ln(x)\cdot 1 \ dx = x\cdot\ln(x)-\int x\cdot\dfrac{1}{x} \ dx=$$$ $$$=x\cdot\ln(x)-\int 1 \ dx=x\cdot\ln(x)-x+C $$$
When we have to integrate logarithms it is often useful to take $$u(x)=\ln (x)$$ since its derivative can generally be simplified with other terms in the integral.
Solution:
$$\displaystyle\int\ln(x) \ dx=x\cdot\ln(x)-x+C $$