How do you find the indefinite integral of int (3/x)dx(3x)dx?

1 Answer
Nov 11, 2016

int(3/x)dx=3lnx +C = ln(Ax^3) (3x)dx=3lnx+C=ln(Ax3)

Explanation:

You should remember a standard special case:

d/dxlnx = 1/x <=> int 1/xdx = lnx + C ddxlnx=1x1xdx=lnx+C

Hence, int(3/x)dx=3lnx +C (3x)dx=3lnx+C

NB If we write C=lnAC=lnA we can also write the solution as

int(3/x)dx=3lnx + lnA (3x)dx=3lnx+lnA
:. int(3/x)dx=lnx^3 + lnA
:. int(3/x)dx=ln(Ax^3)