Integrate ln(x)

Our A-level knowledge does not give us any identities to be able to integrate this from memory. But what we do have from memory is that the derivative of ln(x) is 1/x. Why would this help us? Let's take a look at integration by parts.∫uv' dx = uv - ∫ vu' dx where the "dash" just represents the derivative i.e u' is du/dx and v' is dv/dx.We have to assign values to u and v'. When we multiply anything by 1 we get the same value as we started with so we will use this property to help us answer the question. Let's set u = ln(x) and v' = 1 u = ln(x) u'= ? v= ? v'= 1To obtain v, we must integrate v' which gives us a value of x. Similarly, we must differentiate u to obtain u' which, from memory, is 1/x.u = ln(x) u'= 1/x v= x v'= 1Substituting this into our equation about we get: ∫ln(x) dx = xln(x) - ∫x/x dx = xln(x) - ∫1 dxTherefore our final answer is: xln(x) -x + c Note: Don't forget the constant in indefinite integration!