Biết I = intlimits_0^4 {xln (2x + 1)dx} = frac{a}{b}ln 3 - c, trong đó a, b, c là các số nguyên dương và frac{b}{c} là phân số tối giản

Đặt \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2}}}{2} \end{array} \right. \Rightarrow I = \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\frac{{{x^2}}}{{2x + 1}}dx}\) \(\Rightarrow I = \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\left( {\frac{x}{2} - \frac{1}{4} + \frac{1}{{4(2x + 1)}}} \right)dx}\)

\(= \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \left. {\left( {\frac{{{x^2}}}{4} - \frac{1}{4}x + \frac{1}{8}\ln (2x + 1)} \right)} \right|_0^4\)

\(\Rightarrow I = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right. \Rightarrow S = a + b + c = 70.\)

Cách khác: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)

Đặt: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)

\(\Rightarrow I = \left. {\left[ {\frac{{4{x^2} - 1}}{8}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\frac{{2x - 1}}{4}dx}\)

\(\Rightarrow I = \frac{{63}}{8}\ln 9 - \left. {\frac{{({x^2} - x)}}{4}} \right|_0^4 = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right.\)

\(\Rightarrow S = a + b + c = 70.\)