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
