Question

Question #fefc1

Answer 1

`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)))`

Explanation:

`I = int udv = uv - int vdu`

`u=ln(3x-x^2) => du=(3-2x)/(3x-x^2)dx`

`dv = xdx => v = int xdx = x^2/2`

`I = x^2/2ln(3x-x^2) - int x^2/2 (3-2x)/(3x-x^2)dx`

`I = 1/2[x^2ln(3x-x^2) - int (x^2(2x-3))/(x^2-3x)dx]`

`I_1 = int (2x^3-3x^2)/(x^2-3x)dx = int (2x^3-6x^2+3x^2-9x+9x)/(x^2-3x)dx`

`I_1 = int (2x(x^2-3x)+3(x^2-3x)+9x)/(x^2-3x)dx`

`I_1 = int (2x+3+(9x)/(x(x-3)))dx = int (2x+3+9/(x-3))dx`

`I_1 = x^2 + 3x + 9ln|x-3|+C`

`I = 1/2[x^2ln(3x-x^2) - x^2-3x-9ln|x-3|]|_a^(a^2)`

`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 = 1/2(3a-2a^2-a^4+9ln|a-3|/|a^2-3|+a^2ln((3a^2-a^4)^(a^2)/(3a-a^2)))`