Đáp án:

Giải thích các bước giải: c) \(I = \int\limits_0^1 {\ln (2x + 1)dx} \) Đặt \(\left\{ \begin{array}{l}u = \ln \left( {2x + 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \frac{2}{{2x + 1}}dx\\v = x\end{array} \right.\) \( \Rightarrow I = \left. {x\ln \left( {2x + 1} \right)} \right|_0^1 - \int\limits_0^1 {\frac{{2x}}{{2x + 1}}dx} \) \( = \ln 3 - \int\limits_0^1 {\left( {1 - \frac{1}{{2x + 1}}} \right)dx} \) \( = \ln 3 - \left. {\left( {x - \frac{{\ln \left( {2x + 1} \right)}}{2}} \right)} \right|_0^1\) \( = \ln 3 - \left( {1 - \frac{{\ln 3}}{2}} \right) = \frac{3}{2}\ln 3 - 1\) d) \(I = \int\limits_2^3 {\left[ {\ln \left( {x - 1} \right) - \ln \left( {x + 1} \right)} \right]dx} \) \( = \int\limits_2^3 {\ln \left( {x - 1} \right)dx} - \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \) \( = J - K\) với \(J = \int\limits_2^3 {\ln \left( {x - 1} \right)dx} \) và \(K = \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \). +) Tính \(J = \int\limits_2^3 {\ln \left( {x - 1} \right)dx} \). Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x - 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{{x - 1}}\\v = x\end{array} \right.\) \( \Rightarrow J = \left. {x\ln \left( {x - 1} \right)} \right|_2^3 - \int\limits_2^3 {\frac{x}{{x - 1}}dx} \) \( = 3\ln 2 - \int\limits_2^3 {\left( {1 + \frac{1}{{x - 1}}} \right)dx} \) \( = 3\ln 2 - \left. {\left( {x + \ln \left( {x - 1} \right)} \right)} \right|_2^3\) \( = 3\ln 2 - 3 - \ln 2 + 2\) \( = 2\ln 2 - 1\). +) Tính \(K = \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \). Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{{x + 1}}\\v = x\end{array} \right.\) \( \Rightarrow K = \left. {x\ln \left( {x + 1} \right)} \right|_2^3 - \int\limits_2^3 {\frac{x}{{x + 1}}dx} \) \( = 3\ln 4 - 2\ln 3 - \int\limits_2^3 {\left( {1 - \frac{1}{{x + 1}}} \right)dx} \) \( = 6\ln 2 - 2\ln 3 - \left. {\left( {x - \ln \left( {x + 1} \right)} \right)} \right|_2^3\) \( = 6\ln 2 - 2\ln 3 - 3 + \ln 4 + 2 - \ln 3\) \( = 8\ln 2 - 3\ln 3 - 1\). \( \Rightarrow I = J - K\) \( = 2\ln 2 - 1 - \left( {8\ln 2 - 3\ln 3 - 1} \right)\) \( = 3\ln 3 - 6\ln 2\) e) \(I = \int\limits_{\frac{1}{2}}^2 {\left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx} \)\( = \int\limits_{\frac{1}{2}}^2 {{e^{x + \frac{1}{x}}}} dx + \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx} \) \( = J + K\) với \(J = \int\limits_{\frac{1}{2}}^2 {{e^{x + \frac{1}{x}}}} dx\) và \(K = \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx} \) +) Tính \(J = \int\limits_{\frac{1}{2}}^2 {{e^{x + \frac{1}{x}}}} dx\) Đặt \(\left\{ \begin{array}{l}u = {e^{x + \frac{1}{x}}}\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \left( {1 - \frac{1}{{{x^2}}}} \right){e^{x + \frac{1}{x}}}dx\\v = x\end{array} \right.\) \( \Rightarrow J = \left. {x{e^{x + \frac{1}{x}}}} \right|_{\frac{1}{2}}^2 - \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx} \) \( = \left. {x{e^{x + \frac{1}{x}}}} \right|_{\frac{1}{2}}^2 - K\) \( = 2{e^{\frac{5}{2}}} - \frac{1}{2}{e^{\frac{5}{2}}} - K = \frac{3}{2}{e^{\frac{5}{2}}} - K\) Suy ra \(I = J + K\) \( = \frac{3}{2}{e^{\frac{5}{2}}} - K + K = \frac{3}{2}{e^{\frac{5}{2}}}\). g) \(I = \int\limits_0^{\frac{\pi }{2}} {x\cos x{{\sin }^2}xdx} \) Đặt \(u = x,dv = \cos x{\sin ^2}xdx\) \( \Rightarrow du = dx\). Ta tìm \(v = \int {\cos x{{\sin }^2}xdx} \). Đặt \(\sin x = t \Rightarrow dt = \cos xdx\) \( \Rightarrow \int {\cos x{{\sin }^2}xdx} = \int {{t^2}dt} \) \( = \frac{{{t^3}}}{3} + C = \frac{{{{\sin }^3}x}}{3} + C\) Chọn \(v = \frac{{{{\sin }^3}x}}{3}\) ta có: \(I = \int\limits_0^{\frac{\pi }{2}} {x\cos x{{\sin }^2}xdx} \)\( = \left. {\frac{{x{{\sin }^3}x}}{3}} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}x}}{3}dx} \) \( = \frac{\pi }{6} - \frac{1}{3}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right)\sin xdx} \) \( = \frac{\pi }{6} - \frac{1}{3}J\) Đặt \(\cos x = t \Rightarrow dt = - \sin xdx\) \( \Rightarrow J = \int\limits_1^0 {\left( {1 - {t^2}} \right).\left( { - dt} \right)} \) \( = \int\limits_0^1 {\left( {1 - {t^2}} \right)dt} \) \( = \left. {\left( {t - \frac{{{t^3}}}{3}} \right)} \right|_0^1 = \frac{2}{3}\) Vậy \(I = \frac{\pi }{6} - \frac{1}{3}J\) \( = \frac{\pi }{6} - \frac{1}{3}.\frac{2}{3} = \frac{\pi }{6} - \frac{2}{9}\).