Answer
$$\frac{1}{2}{\tan ^{ - 1}}\left( {2{{\sin }^2}x - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx \cr
& {\text{Rewrite}} \cr
& = \int {\frac{{\sin x\cos x}}{{{{\sin }^4}x + {{\left( {{{\cos }^2}x} \right)}^2}}}} dx \cr
& {\text{Use pythagorean identities}} \cr
& = \int {\frac{{\sin x\cos x}}{{{{\sin }^4}x + {{\left( {1 - {{\sin }^2}x} \right)}^2}}}} dx \cr
& = \int {\frac{{\sin x\cos x}}{{{{\sin }^4}x + 1 - 2{{\sin }^2}x + {{\sin }^4}x}}} dx \cr
& = \int {\frac{{\sin x\cos x}}{{2{{\sin }^4}x - 2{{\sin }^2}x + 1}}} dx \cr
& = \int {\frac{{\sin x\cos x}}{{2{{\left( {{{\sin }^2}x} \right)}^2} - 2{{\sin }^2}x + 1}}} dx \cr
& {\text{Let }}u = {\sin ^2}x \to du = 2\sin x\cos xdx \cr
& {\text{Substituting}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{2{u^2} - 2u + 1}}} \cr
& = \frac{1}{4}\int {\frac{{du}}{{{u^2} - u + 1/2}}} \cr
& {\text{Completing the square}} \cr
& = \frac{1}{4}\int {\frac{{du}}{{{u^2} - u + 1/4 + 1/2 - 1/4}}} \cr
& = \frac{1}{4}\int {\frac{{du}}{{{{\left( {u - 1/2} \right)}^2} + 1/4}}} \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\left( {\frac{1}{{1/2}}{{\tan }^{ - 1}}\left( {\frac{{u - 1/2}}{{1/2}}} \right)} \right) + C \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {2u - 1} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {2{{\sin }^2}x - 1} \right) + C \cr} $$
