Answer
$A \approx 1.6925$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = x\ln \left( {x + 1} \right){\text{ and }}y = 3x - {x^2} \cr
& {\text{From the graph we have the intersection points}} \cr
& x = 0{\text{ and }}x \approx 1.93 \cr
& {\text{The area of the region bounded by the curves is given by:}} \cr
& A = \int_0^{1.93} {\left[ {\left( {3x - {x^2}} \right) - x\ln \left( {x + 1} \right)} \right]} dx \cr
& \cr
& {\text{* Integrating }}\int {x\ln \left( {x + 1} \right)} dx \cr
& u = \ln \left( {x + 1} \right),{\text{ }}du = \frac{1}{{x + 1}}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \underbrace {\frac{1}{2}\int {\frac{{{x^2}}}{{x + 1}}} dx}_{{\text{Use the long division}}} \cr
& = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\int {\left( {x - 1 + \frac{1}{{x + 1}}} \right)} dx \cr
& = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right) + C \cr
& = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{{{x^2}}}{4} + \frac{1}{2}x - \frac{1}{2}\ln \left| {x + 1} \right| + C \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A \approx \int_0^{1.93} {\left[ {\left( {3x - {x^2}} \right) - x\ln \left( {x + 1} \right)} \right]} dx}_ \Downarrow \cr
& A \approx \left[ {\frac{3}{2}{x^2} - \frac{1}{3}{x^3} - \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) + \frac{{{x^2}}}{4} - \frac{1}{2}x + \frac{1}{2}\ln \left| {x + 1} \right|} \right]_0^{1.93} \cr
& {\text{Evaluating the limits we obtain}} \cr
& A \approx \left[ {1.6925} \right] - \left[ 0 \right] \cr
& A \approx 1.6925 \cr} $$
