Let J=∫ln(x)sin(3x)dx; let u=ln(x), du=dx/x; dv=sin(3x)dx, v=-⅓cos(3x).
∫udv=-⅓ln(x)cos(3x)+∫(cos(3x)/(3x)dx.
∫(cos(3x)/(3x))dx is a non-elementary integral represented by Ci(3x). Ci stands for Cosine Integral.
So J=-⅓ln(x)cos(3x)+Ci(3x)+A where A is a constant.
The integral can be resolved by considering the power series for cosine:
cos(3x)=1-(3x)2/2!+(3x)4/4!-...+(-1)n(3x)2n/(2n)!+...;
cos(3x)/(3x)=1/(3x)-3x/2!+(3x)3/4!-...+(-1)n(3x)2n-1/(2n)!+...
On integration we get: ⅓ln(x)-3x2/4+9x4/32-...+(-1)n(32n-1)x2n/((2n)!(2n)).
J=A-⅓ln(x)cos(3x)+⅓ln(x)+∑(-1)n(32n-1)x2n/((2n)!(2n)) for integer 0<n<∞.