Hope the question is of the form
#int1/(sin^4x+cos^4x+sin^2xcos^2x)dx#
Let
#u=cos^2x, v=sin^2x#
#u^3-v^3=(u-v)(u^2+uv+v^2)#
Thus
#u^2+uv+v^2=(u^3-v^3)/(u-v)#
#1/(u^2+uv+v^2)=(u-v)/(u^3-v^3)#
Substituting
#1/(sin^4x+cos^4x+sin^2xcos^2x)=(cos^2x-sin^2x)/((cos^2x)^3-(sin^2x)^3)#
Dividing throughout by #cos^6x#
#1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^4x-sec^2xtan^2x)/(1-tan^6x)#
#1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^2x(sec^2x-tan^2x))/(1-tan^6x)#
#sec^2x-tan^2x=1#
#(sec^2x)/(1-tan^6x)#
#"Now, ..."#
#int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=int(sec^2x)/(1-tan^6x)dx#
Thus,
If
#t=tanx#
#dt=sec^2xdx#
#int(sec^2x)/(1-tan^6x)dx=int(dt)/(1-t^6)dx#
#1/(1-t^6)=A/(1-t^3)+B/(1+t^3)#
#1=A(1+t^3)+B(1-t^3)#
#1=A+At^3+B-Bt^3#
#1=(A+B)+(A-B)t^3#
Equating the coefficients of like powers of t
#1=A+B#
#0=A-B#
IE
#A=B#
#A=1/2, B=1/2#
#1/(1-t^6)=A/(1-t^3)+B/(1+t^3)#
#1/(1-t^6)=(1/2)/(1-t^3)-(1/2)/(1+t^3)#
#1/(1-t^6)=1/2(1/(1-t^3)-1/(1+t^3))#
Let
# I_1=int1/(1-t^3)dt#
#I_2=int1/(1+t^3)dt#
#1/(1-t^3)=1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)#
#1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)#
#1=C(1+t+t^2)+(Dt+E)(1-t)#
#1=C+Ct+Ct^2+Dt+E-Dt^2-Et#
#1=(C+E)+(C+D-E)t+(C-D)t^2#
Equating the coefficients of like powers of t
#0=C+E#
#E=-C#
#0=C+D-E#
#0=C+D+C#
#0=2C+D#
#D=-2C#
#C-D=1#
#C+2C=1#
#3C=1#
#C=1/3#
#D=-2/3#
#E=-1/3#
#1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)#
#1/((1-t)(1+t+t^2))=(1/3)/(1-t)+(-2/3t-1/3)/(1+t+t^2)#
#1/((1-t)(1+t+t^2))=1/3(1/(1-t)-(2t+1)/(1+t+t^2))#
#1/((1-t)(1+t+t^2))=1/3(-1/(t-1)-(2t+1)/(t^2+t+1))#
#1/((1-t)(1+t+t^2))=-1/3(1/(t-1)+(2t+1)/(t^2+t+1))#
#I_1=int1/(1+t^3)dt#
#I_1=int-1/3(1/(t-1)+(2t+1)/(t^2+t+1))dt#
#I_1=-int1/3(1/(t-1)+(2t+1)/(t^2+t+1))dt#
#I_1=-1/3(ln(t-1)+ln(t^2+t+1))#
#-1/3ln(t-1)(t^2+t+1)#
#I_1=-1/3ln(t^3-1)#
#1/(1+t^3)=1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)#
#1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)#
#1=C(1-t+t^2)+(Dt+E)(1+t)#
#1=C-Ct+Ct^2+Dt+E+Dt^2+Et#
#1=(C+E)+(-C+D-E)t+(C+D)t^2#
Equating the coefficients of like powers of t
#0=C+E#
#E=-C#
#0=-C+D-E#
#0=D#
#D=0#
#C+D=1#
#C+0=1#
#C=1#
#D=-2/3#
#E=-1#
#1/((1+t)(1-t+t^2))=1/(1+t)+(0t-1)/(1-t+t^2)#
#1/((1+t)(1-t+t^2))=1/(1+t)-1/(1-t+t^2)#
#I_2=int1/(1+t^3)dt#
#I_2=int(1/(1+t)-1/(1-t+t^2))dt#
#int(1/(1+t)dt=ln(t+1)#
#int(1/(1-t+t^2))dt=int(1/(t^2-t+1))dt#
Completing the squares
#t^2-t+1=t^2-2*1/2*t+(1/2)^2+1-(1/2)^2#
#(t-1/2)^2+3/4#
#(t-1/2)^2+(sqrt3/2)^2#
Thus,
#int(1/(t^2-t+1))dt=int1/((t-1/2)^2+(sqrt3/2)^2)dt#
#int1/((t-1/2)^2+(sqrt3/2)^2)dt=1/(sqrt3/2)tan^-1(t-1/2)#
#int1/((t-1/2)^2+(sqrt3/2)^2)dt=(2/sqrt3)tan^-1(t-1/2)#
#I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)#
#t=tanx#
#I_1=-1/3ln(t^3-1)#
#I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)#
#I=1/2(I_1-I_2)#
#I=1/2(-1/3ln(t^3-1)-(ln(t+1)+(2/sqrt3)tan^-1(t-1/2)))#
#I=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)#
Thus,
#int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)#
#t=tanx#
#int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(tan^3x-1)-ln(tanx+1)-2/sqrt3tan^-1(tanx-1/2)+C#
