\[\begin{array}{l} 3.13.\,\,\,\\ I = \int {\left( {x + 1} \right)\sin xdx = } \int {x\sin xdx} + \int {\sin xdx} \\ = \int {x\sin xdx} - \cos x + C.\\ Tinh\,\,\,J = \int {x\sin xdx} \\ Dat\,\,\left\{ \begin{array}{l} u = x\\ dv = {\mathop{\rm sinx}\nolimits} dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = - \cos x \end{array} \right.\\ \Rightarrow J = - x\cos x + \int {\cos xdx} = - x\cos x + \sin x\\ \Rightarrow I = - x\cos x + \sin x - \cos x + C = - \left( {x + 1} \right)\cos x + \sin x + C.\\ 3.14.\,\,I = \int {x\ln \left( {x + 1} \right)dx} \\ Dat\,\,\left\{ \begin{array}{l} u = \ln \left( {x + 1} \right)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{{x + 1}}dx\\ v = \frac{{{x^2}}}{2} \end{array} \right.\\ \Rightarrow I = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \int {\frac{{{x^2}}}{{2\left( {x + 1} \right)}}dx} \\ = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\int {\frac{{{x^2} - 1 + 1}}{{x + 1}}dx} \\ = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\int {\left( {x - 1 + \frac{1}{{x + 1}}} \right)dx} \\ = \frac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right) + C.\\ 3.15)\,\,\,I = \int {x\sqrt {x - 1} dx} \\ Dat\,\,\,\sqrt {x - 1} = t \Rightarrow {t^2} = x - 1 \Rightarrow x = {t^2} + 1 \Rightarrow dx = 2tdt\\ \Rightarrow I = \int {\left( {{t^2} + 1} \right)t.2tdt} = \int {\left( {2{t^4} + 2{t^2}} \right)dt} \\ = \frac{{2{t^5}}}{5} + \frac{{2{t^3}}}{3} + C\\ = \frac{2}{5}{\left( {\sqrt {x - 1} } \right)^5} + \frac{2}{3}{\left( {\sqrt {x - 1} } \right)^2} + C. \end{array}\] Em tự chọn đáp án nhé em.