Follow every step. The layout matters: examiners award method marks for clearly stating u and du. Level 1: the classic ∫ 2x(x² + 1)⁵ dx Find ∫ 2x(x² + 1)⁵ dx. Choose u. The nest is x²+1, so let u = x² + 1 . Find du. du/dx = 2x , so du = 2x dx . Substitute. The integrand is (x²+1)⁵ · (2x dx) = u⁵ du. So the integral becomes ∫ u⁵ du . Integrate. ∫u⁵ du = u⁶/6 + c . Substitute back. Replace u with x²+1. ✓ (x² + 1)⁶ / 6 + c Level 2: fixing a constant ∫ x·e^(x²) dx Find ∫ x e^(x²) dx. Let u = x² (the exponent is the nest). du = 2x dx . But we only have x dx , so rearrange: x dx = ½ du . Substitute: ∫ e^u · ½ du = ½ ∫ e^u du . = ½ e^u + c . Back-substitute u = x². ✓ ½ e^(x²) + c Level 2: linear inside ∫ cos(3x + 1) dx Find ∫ cos(3x + 1) dx. Let u = 3x + 1 . du = 3 dx ⟹ dx = ⅓ du . ∫ cos u · ⅓ du = ⅓ ∫ cos u du = ⅓ sin u + c . Back-substitute. ✓ ⅓ sin(3x + 1) + c, (this is the “divide by the coefficient” shortcut, and this is WHY it works) Level 3: producing a logarithm ∫ (2x + 3)/(x² + 3x) dx Find ∫ (2x + 3)/(x² + 3x) dx. For a fraction, try the denominator: u = x² + 3x . du = (2x + 3) dx . The numerator is exactly du. Perfect. ∫ (1/u) du = ln|u| + c . Back-substitute. ✓ ln|x² + 3x| + c Level 3: trig power ∫ sin⁴x · cos x dx Find ∫ sin⁴x cos x dx. Let u = sin x (it’s raised to a power; and its derivative cos x is sitting right there). du = cos x dx . ∫ u⁴ du = u⁵/5 + c . Back-substitute. ✓ sin⁵x / 5 + c Level 4: the stubborn extra x ∫ x√(x − 2) dx Find ∫ x√(x − 2) dx. (Here du does NOT clear all the x’s, you must rearrange.) Let u = x − 2 , so du = dx . Substituting gives ∫ x·√u du , but there’s still an x! Not allowed. Fix: rearrange the substitution to express x in terms of u: x = u + 2 . Now: ∫ (u + 2)·u^(1/2) du = ∫ (u^(3/2) + 2u^(1/2)) du . Integrate: = (2/5)u^(5/2) + (4/3)u^(3/2) + c . Back-substitute u = x − 2. ✓ (2/5)(x−2)^(5/2) + (4/3)(x−2)^(3/2) + c