Question #fefc1

1 Answer
Apr 24, 2017

I = 1/2(3a-2a^2-a^4+9ln|a-3|/|a^2-3|+a^2ln((3a^2-a^4)^(a^2)/(3a-a^2)))I=123a2a2a4+9ln|a3||a23|+a2ln(3a2a4)a23aa2

Explanation:

I = int udv = uv - int vduI=udv=uvvdu

u=ln(3x-x^2) => du=(3-2x)/(3x-x^2)dxu=ln(3xx2)du=32x3xx2dx

dv = xdx => v = int xdx = x^2/2dv=xdxv=xdx=x22

I = x^2/2ln(3x-x^2) - int x^2/2 (3-2x)/(3x-x^2)dxI=x22ln(3xx2)x2232x3xx2dx

I = 1/2[x^2ln(3x-x^2) - int (x^2(2x-3))/(x^2-3x)dx]I=12[x2ln(3xx2)x2(2x3)x23xdx]

I_1 = int (2x^3-3x^2)/(x^2-3x)dx = int (2x^3-6x^2+3x^2-9x+9x)/(x^2-3x)dxI1=2x33x2x23xdx=2x36x2+3x29x+9xx23xdx

I_1 = int (2x(x^2-3x)+3(x^2-3x)+9x)/(x^2-3x)dxI1=2x(x23x)+3(x23x)+9xx23xdx

I_1 = int (2x+3+(9x)/(x(x-3)))dx = int (2x+3+9/(x-3))dxI1=(2x+3+9xx(x3))dx=(2x+3+9x3)dx

I_1 = x^2 + 3x + 9ln|x-3|+CI1=x2+3x+9ln|x3|+C

I = 1/2[x^2ln(3x-x^2) - x^2-3x-9ln|x-3|]|_a^(a^2)I=12[x2ln(3xx2)x23x9ln|x3|]a2a

I = 1/2(a^4ln(3a^2-a^4)-a^4-3a^2-9ln|a^2-3|-a^2ln(3a-a^2)+a^2+3a+9ln|a-3|)I=12(a4ln(3a2a4)a43a29lna23a2ln(3aa2)+a2+3a+9ln|a3|)

I = 1/2(3a-2a^2-a^4+9ln|a-3|/|a^2-3|+a^2ln((3a^2-a^4)^(a^2)/(3a-a^2)))I=123a2a2a4+9ln|a3||a23|+a2ln(3a2a4)a23aa2