How to solve: ∫x ln (1-3x) dx?

Answer 1

Use integration by parts.

Answer 2

234

Answer 3

♠ Y=∫dx*x*ln(1-3x); by parts:
u= ln(1-3x), du=-3dx/(1-3x);
dv=dx*x; v=0.5*x^2;
♣ Y=0.5x^2* ln(1-3x) -∫(-3dx/(1-3x))*(0.5x^2) =
= 0.5x^2* ln(1-3x) +1.5∫dx*x^2/(1-3x) = Y1 + Y2;
♦ Y2=1.5∫dx*x^2/(1-3x) = -0.5∫dx*x – (1/6)∫dx +(1/6)∫dx/(1-3x) =
=-0.25x^2 –(1/6)x –(1/6)ln(1-3x)/3;
♥ Y=(0.5x^2 –1/18)* ln(1-3x) -0.25x^2 –(1/6)x +C;