a) Đặt \(u = x + \ln x;dv = {x^2}dx\) \( \Rightarrow \left\{ \begin{array}{l}du = \left( {1 + \dfrac{1}{x}} \right)dx\\v = \dfrac{{{x^3}}}{3}\end{array} \right.\)
Khi đó \(\int {(x + \ln x){x^2}dx} \)\( = \dfrac{{{x^3}}}{3}\left( {x + \ln x} \right) - \int {\dfrac{{{x^3}}}{3}\left( {1 + \dfrac{1}{x}} \right)dx} \)
\( = \dfrac{{{x^4}}}{3} + \dfrac{{{x^3}}}{3}\ln x - \int {\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{3}} \right)dx} \) \( = \dfrac{{{x^4}}}{3} + \dfrac{{{x^3}}}{3}\ln x - \dfrac{{{x^4}}}{{12}} - \dfrac{{{x^3}}}{9} + C\) \( = \dfrac{{{x^4}}}{4} + \dfrac{{{x^3}}}{3}\left( {\ln x - \dfrac{1}{3}} \right) + C\).
b) Đặt \(u = x + {\sin ^2}x,dv = \sin xdx\) \( \Rightarrow \left\{ \begin{array}{l}du = \left( {1 + 2\sin x\cos x} \right)dx\\v = - \cos x\end{array} \right.\)
c) Đặt \(u = x + {e^x},dv = {e^{2x}}dx\) \( \Rightarrow \left\{ \begin{array}{l}du = \left( {1 + {e^x}} \right)dx\\v = \dfrac{{{e^{2x}}}}{2}\end{array} \right.\)
d) Đặt \(\left\{ \begin{array}{l}u = x + \sin x\\dv = \dfrac{1}{{{{\cos }^2}x}}dx\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}du = \left( {1 + \cos x} \right)dx\\v = \tan x\end{array} \right.\)
