Find the value of \(\int_0^1 {t\ln (t + 1){\text{d}}t} \).

EITHER

attempt at integration by substitution     (M1)

using \(u = t + 1{\text{, d}}u = {\text{d}}t\), the integral becomes

\(\int_1^2 {(u - 1)\ln u{\text{d}}u} \)     A1

then using integration by parts     M1

\(\int_1^2 {(u - 1)\ln u{\text{d}}u}  = \left[ {\left( {\frac{{{u^2}}}{2} - u} \right)\ln u} \right]_1^2 - \int_1^2 {\left( {\frac{{{u^2}}}{2} - u} \right) \times \frac{1}{u}{\text{d}}u} \)     A1

\( =  - \left[ {\frac{{{u^2}}}{4} - u} \right]_1^2\)     (A1)

\( = \frac{1}{4}\,\,\,\,\,\)(accept 0.25)     A1

OR

attempt to integrate by parts     (M1)

correct choice of variables to integrate and differentiate     M1

\(\int_0^1 {t\ln (t + 1){\text{d}}t}  = \left[ {\frac{{{t^2}}}{2}\ln (t + 1)} \right]_0^1 - \int_0^1 {\frac{{{t^2}}}{2} \times \frac{1}{{t + 1}}{\text{d}}t} \)     A1

\( = \left[ {\frac{{{t^2}}}{2}\ln (t + 1)} \right]_0^1 - \frac{1}{2}\int_0^1 {t - 1 + \frac{1}{{t + 1}}{\text{d}}t} \)     A1

\( = \left[ {\frac{{{t^2}}}{2}\ln (t + 1)} \right]_0^1 - \frac{1}{2}\left[ {\frac{{{t^2}}}{2} - t + \ln (t + 1)} \right]_0^1\)     (A1)

\( = \frac{1}{4}\,\,\,\,\,\)(accept 0.25)     A1

[6 marks]

Again very few candidates gained full marks on this question. The most common approach was to begin by integrating by parts, which was done correctly, but very few candidates then knew how to integrate \(\frac{{{t^2}}}{{t + 1}}\). Those who began with a substitution often made more progress. Again a number of candidates were let down by their inability to simplify appropriately.