Integration by parts

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.