Concept: Factor denominator using sum of cubes, substitute \(t = \sin x - \cos x\), express everything in t, use partial fractions, and integrate.
Step 1: \(\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)\). Let \(t = \sin x - \cos x\). Then \(t^2 = 1 - 2\sin x\cos x \Rightarrow \sin x\cos x = (1-t^2)/2\). Also \((\sin x+\cos x)^2 = 1 + 2\sin x\cos x = 2-t^2\). \(dx = dt/(\cos x+\sin x) = dt/\sqrt{2-t^2}\).
Step 2: Denominator = \(\sqrt{2-t^2} \cdot (1 - (1-t^2)/2) = \sqrt{2-t^2} \cdot (1+t^2)/2\). Integral: \(\int \frac{dx}{\sin^3x+\cos^3x} = \int \frac{dt/\sqrt{2-t^2}}{\sqrt{2-t^2}(1+t^2)/2} = \int \frac{2 dt}{(1+t^2)(2-t^2)}\).
Step 3: Partial fractions: \(\frac{2}{(1+t^2)(2-t^2)} = \frac{2/3}{1+t^2} + \frac{2/3}{2-t^2}\). Integrate: \(\frac23\tan^{-1}t + \frac23 \cdot \frac{1}{2\sqrt2} \log\left|\frac{\sqrt2+t}{\sqrt2-t}\right| + C\). So \(A = \frac{1}{3\sqrt2}, B = \frac23\).
Step 4: \(B/A = \frac{2/3}{1/(3\sqrt2)} = 2\sqrt2\). The argument is \(t = \sin x - \cos x\).
Step 5: Write the final answer. \(\boxed{\left(2\sqrt2,\ \sin x-\cos x\right)}\)