1. We are asked to find the antiderivative of \(4x \cos(2 - 3x)\) using integration by parts if appropriate.\n\n2. Let \(u = 4x\) and \(dv = \cos(2 - 3x) dx\). Then \(du = 4 dx\) and we find \(v = \int \cos(2 - 3x) dx\).\n\n3. To find \(v\), let \(w = 2 - 3x\), so \(dw = -3 dx\), or \(dx = -\frac{1}{3} dw\). Then\n$$v = \int \cos(w) \left(-\frac{1}{3}\right) dw = -\frac{1}{3} \sin(w) + C = -\frac{1}{3} \sin(2 - 3x) + C.$$\n\n4. By integration by parts formula \(\int u \, dv = uv - \int v \, du\), we have\n$$\int 4x \cos(2 - 3x) dx = 4x \left(-\frac{1}{3} \sin(2 - 3x) \right) - \int -\frac{1}{3} \sin(2 - 3x) (4) dx$$\n$$= -\frac{4x}{3} \sin(2 - 3x) + \frac{4}{3} \int \sin(2 - 3x) dx.$$\n\n5. To evaluate \(\int \sin(2 - 3x) dx\), substitute \(w = 2 - 3x\), \(dx = -\frac{1}{3} dw\), so\n$$\int \sin(2 - 3x) dx = \int \sin(w) \left(-\frac{1}{3}\right) dw = -\frac{1}{3} \int \sin(w) dw = -\frac{1}{3} (-\cos(w)) + C = \frac{1}{3} \cos(2 - 3x) + C.$$\n\n6. Substitute back this in step 4 result:\n$$-\frac{4x}{3} \sin(2 - 3x) + \frac{4}{3} \cdot \frac{1}{3} \cos(2 - 3x) + C = -\frac{4x}{3} \sin(2 - 3x) + \frac{4}{9} \cos(2 - 3x) + C.$$\n\n7. Now, evaluate \(\int \frac{x}{\sqrt{1 - x^4}} dx\). Notice we can rewrite \(x^4 = (x^2)^2\), and try substitution \(t = x^2\), so \(dt = 2x dx\) or \(x dx = \frac{1}{2} dt\).\n\n8. Substituting into the integral:\n$$\int \frac{x}{\sqrt{1 - x^4}} dx = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{\sqrt{1 - t^2}} dt.$$\n\n9. The integral \(\int \frac{1}{\sqrt{1 - t^2}} dt = \arcsin t + C\). Therefore,\n$$\frac{1}{2} \int \frac{1}{\sqrt{1 - t^2}} dt = \frac{1}{2} \arcsin t + C = \frac{1}{2} \arcsin(x^2) + C.$$\n\n10. Final answers:\n(i) $$\int 4x \cos(2 - 3x) dx = -\frac{4x}{3} \sin(2 - 3x) + \frac{4}{9} \cos(2 - 3x) + C.$$\n(ii) $$\int \frac{x}{\sqrt{1 - x^4}} dx = \frac{1}{2} \arcsin(x^2) + C.$$