Tính các tích phân sau bằng phương pháp tích phân từng phần:

$$\begin{gathered} a) {\int }_1^{e^4}\sqrt{x}\ln xdx; b) {\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{xdx}{{\sin }^{2}x}; \\ c) {\int }_0^{\pi }(\pi -x)\sin xdx; c) {\int }_{-1}^0(2x+3)e^{-x}dx. \end{gathered}$$

Hướng dẫn trả lời

a) Đặt $\left\{\begin{array}{l}u=\ln x\\ dv=\sqrt{x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=\frac{dx}{x}\\ v=\frac{2}{3}{x}^{\frac{3}{2}}\end{array}\right.\right.$

 ${\int }_1^{e^4}\sqrt{x}\ln xdx=\frac{2}{3}{x}^{\frac{3}{2}}\ln x{|}_1^{e^4}-\frac{2}{3}{\int }_1^{e^4}{x}^{\frac{1}{2}}dx=\frac{2}{3}\sqrt{x^3}\ln x{|}_1^{e^4}-\frac{4}{9}\sqrt{x^3}{|}_1^{e^4}=\frac{4}{9}(5e^6+1)$

b) Đặt $\left\{\begin{array}{l}u=x\\ dv=\frac{1}{{\sin }^{2}x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=-cotx\end{array}\right.\right.$

  ${\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{xdx}{{\sin }^{2}x}=-xcotx{|}_{\frac{\pi }{6}}^{\frac{\pi }{2}}+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{\cos x}{\sin x}dx=-xcotx{|}_{\frac{\pi }{6}}^{\frac{\pi }{2}}+\ln {\left|\sin x\right|}_{\frac{\pi }{6}}^{\frac{\pi }{2}}=\frac{\pi \sqrt{3}}{6}+\ln 2.$

c) Đặt $\left\{\begin{array}{l}u=\pi -x\\ dv=\sin xdx\end{array}\Rightarrow \left\{\begin{array}{l}du=-dx\\ v=-\cos x\end{array}\right.\right.$

   ${\int }_0^{\pi }(\pi -x)\sin xdx=-(\pi -x)\cos x{|}_0^{\pi }-{\int }_0^{\pi }\cos xdx=-(\pi -x)\cos x{|}_0^{\pi }-\sin x{|}_0^{\pi }=\pi$

d) Đặt $\left\{\begin{array}{l}u=2x+3\\ dv=e^{-x}dx\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=-e^{-x}\end{array}\right.\right.$

  ${\int }_{-1}^0(2x+3)e^{-x}dx=-(2x+3)e^{-x}{|}_{-1}^0+2{\int }_{-1}^0{e}^{-x}dx=-(2x+3)e^{-x}{|}_{-1}^0-2e^{-x}{|}_{-1}^0=3e-5$