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:

$$\displaystyle\int\ln(x) \ dx=x\cdot\ln(x)-x+C $$