Find the partial fractions.
#(3x^2+9x-4)/((x-1)(x^2+4x-1)) = A/(x-1)+(Bx+C)/(x^2+4x-1)#
#(3x^2+9x-4) = A(x^2+4x-1)+(Bx+C)(x-1)#
Eliminate B and C by letting x = 1:
#(3(1)^2+9(1)-4) = A((1)^2+4(1)-1)#
#8=A4#
#A = 2#
#(3x^2+9x-4) = 2(x^2+4x-1)+(Bx+C)(x-1)#
Eliminate B by letting x = 0:
#(3(0)^2+9(0)-4) = 2((0)^2+4(0)-1)+C((0)-1)#
#-4=-2-C#
#C = 2#
#(3x^2+9x-4) = 2(x^2+4x-1)+(Bx+2)(x-1)#
Let x = -1:
#(3(-1)^2+9(-1)-4) = 2((-1)^2+4(-1)-1)+(B(-1)+2)(-1-1)#
#-10=-8+2B-4#
#B = 1#
#int(3x^2+9x-4)/((x-1)(x^2+4x-1))dx = 2int1/(x-1)dx+int(x+2)/(x^2+4x-1)dx#
Multiply the second integral by 1/2 so that the numerator can be multiplied by 2:
#int(3x^2+9x-4)/((x-1)(x^2+4x-1))dx = 2int1/(x-1)dx+1/2int(2x+4)/(x^2+4x-1)dx#
Both integrals become the natural logarithm:
#int(3x^2+9x-4)/((x-1)(x^2+4x-1))dx = 2ln|x-1|+1/2ln|x^2+4x-1|+C#
