Integration by substitution

The idea

• It reverses the chain rule: if the integrand contains a function of g(x) together with a multiple of g′(x), put u = g(x) and du = g′(x) dx.
• Steps: ① choose u (usually the inner function, the denominator, or the expression under a root) ② compute du and solve for dx ③ rewrite the whole integral in u ④ integrate ⑤ return to x.
• For a definite integral, either convert the limits to u or substitute back before applying the original limits.

Two useful standard results

• ∫ f′(x)/f(x) dx = ln|f(x)| + C.
• ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + C, n ≠ −1.

Worked examples

1. ∫ 2x(x² + 1)⁵ dx: u = x² + 1, du = 2x dx → ∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C.
2. ∫ cos⁴x sin x dx: u = cos x, du = −sin x dx → −∫ u⁴ du = −cos⁵x/5 + C.
3. ∫ dx/(x ln x): u = ln x, du = dx/x → ∫ du/u = ln|ln x| + C.
4. ∫ √(2x + 3) dx: u = 2x + 3, du = 2 dx → ½·(2/3)u^3/2 = (2x + 3)^3/2/3 + C.