Hello,
Answer : x ln(3x) - x + ctexln(3x)−x+cte.
Remark that
d/dx (X ln(X) - X) = 1.ln(X) + X/X - 1 = ln(X)ddx(Xln(X)−X)=1.ln(X)+XX−1=ln(X),
therefore
int ln(X) dX = X ln(X) - X + cte∫ln(X)dX=Xln(X)−X+cte
Now, it's easy because, ln(3x) = ln(3) + ln(x)ln(3x)=ln(3)+ln(x), therefore,
int ln(3x) dx = ln(3)x + x ln(x) - x + cte = x ln(3x) - x + cte∫ln(3x)dx=ln(3)x+xln(x)−x+cte=xln(3x)−x+cte
PS. There is a classical trick to find int ln(x) dx∫ln(x)dx if you don't have any idea to test x ln(x) -xxln(x)−x : an integration by parts. The general formula is
int u'v = uv - int uv' (because (uv)' = u'v + uv').
Now, write
int 1 \cdot ln(x) dx = \int u' v, with u=x and v=ln(x). You get
int ln(x) dx = x ln(x) - \int x/x dx = x ln(x) -x + cte
