Standard +0.3 This is a straightforward integration by substitution question with clear guidance (substitution given explicitly). Students must change variables, adjust limits, simplify the integrand algebraically, integrate standard forms, and verify the given answer. While it requires multiple steps and careful algebra, it's a standard C3 exercise with no novel insight required, making it slightly easier than average.
Substituting \(\frac{x}{2x+1} = \frac{u-1}{2u}\) o.e.; converting limits; dividing through by u: \(\frac{1}{4}[u - \ln u]\) o.e. – ft their \(1/4\) (only); must be some evidence of substitution
$\int_0^1 \frac{x}{2x+1}dx$, let $u = 2x + 1 \Rightarrow du = 2dx, x = \frac{u-1}{2}$
When $x = 0, u = 1$; when $x = 1, u = 3$
$= \int_1^3 \frac{(u-1)}{2u} \cdot \frac{1}{2}du = \frac{1}{4}\int_1^3 \frac{u-1}{u}du = \frac{1}{4}\int_1^3 (1 - \frac{1}{u})du = \frac{1}{4}[u - \ln u]_1^3 = \frac{1}{4}[3 - \ln 3 - 1 + \ln 1] = 1/4(2 - \ln 3)$ | M1, A1, B1, M1, A1, E1 [6] | Substituting $\frac{x}{2x+1} = \frac{u-1}{2u}$ o.e.; converting limits; dividing through by u: $\frac{1}{4}[u - \ln u]$ o.e. – ft their $1/4$ (only); must be some evidence of substitution