A-Level Techniques of integration: worked solution

Question

With I=∫P(x)/(1-√x) dx on 0<x<1, use the substitution u=1-√x to find I in the case P(x)=1.

With P(x) = 1 the integral is I = ∫ 1/(1-√x) dx, and the substitution is u = 1-√x. Everything - the integrand AND the dx - must be written in terms of u. From u = 1 - √x, rearrange: √x = 1 - u, and squaring, x = (1-u)². Differentiate to convert dx: dx/du = 2(1-u) × (-1) = -2(1-u) so dx = -2(1-u) du (Equivalently du/dx = -1/(2√x), giving dx = -2√x du = -2(1-u) du.) Substitute both pieces: I = ∫ (1/u) × (-2(1-u)) du = -2 ∫ (1-u)/u du The fraction (1-u)/u cannot be integrated as it stands, so split it term by term: (1-u)/u = 1/u - u/u = 1/u - 1 I = -2 ∫ (1/u - 1) du Integrate each term, using ∫ 1/u du = ln|u|: I = -2( ln|u| - u ) + c = -2 ln|u| + 2u + c Now substitute u = 1 - √x back: I = -2 ln|1-√x| + 2(1-√x) + c = -2 ln|1-√x| + 2 - 2√x + c The constant 2 is absorbed into the arbitrary constant, so I = -2√x - 2 ln|1-√x| + c (On 0 < x < 1 we have 0 < 1-√x < 1, so the modulus is not strictly needed, but keeping it is safe.)