Given: #int (-2x^3-x)/(-4x^2+2x+3) dx#
Multiply the integral by #(-1)/-1#:
#int (2x^3 + x)/(4x^2-2x-3)dx#
Perform the implied division:
#color(white)( (4x^2-2x-3)/color(black)(4x^2-2x-3))(1/2x+1/4color(white)(x+0))/(")" color(white)(x)2x^3+ 0x^2 + x+0)#
#color(white)(".......................")ul(-2x^3+x^2+3/2x)#
#color(white)(".....................................")x^2+5/2x#
#color(white)("..................................")ul(-x^2+1/2x+3/4)#
#color(white)("...............................................")3x+3/4#
#int 1/2x+ 1/4+ (3x+3/4)/(4x^2-2x-3)dx#
Separate into 3 integrals:
#1/2intxdx+ 1/4intdx+ 3int(x+1/4)/(4x^2-2x-3)dx#
Integrate the first two integrals:
#1/4x^2+ 1/4x+ 3int(x+1/4)/(4x^2-2x-3)dx#
Multiply the remaining integral by #8/8#:
#1/4x^2+ 1/4x+ 3/8int(8x+2)/(4x^2-2x-3)dx#
Add to the numerator in the form of #-2+2#:
#1/4x^2+ 1/4x+ 3/8int(8x-2+4)/(4x^2-2x-3)dx#
Separate into two fractions:
#1/4x^2+ 1/4x+ 3/8int(8x-2)/(4x^2-2x-3)+4/(4x^2-2x-3)dx#
Separate into two integrals:
#1/4x^2+ 1/4x+ 3/8int(8x-2)/(4x^2-2x-3)dx+3/8int4/(4x^2-2x-3)dx#
The first integral is the natural logarithm:
#1/4x^2+ 1/4x+ 3/8ln|4x^2-2x-3|+3/2int1/(4x^2-2x-3)dx#
The last integral is done by completing the square:
#1/4x^2+ 1/4x+ 3/8ln|4x^2-2x-3|+(3sqrt13)/52ln|-4x+sqrt13+1|-(3sqrt13)/52ln|4x+sqrt13-1|+ C#
