Bài tập 3.9 trang 165 SBT Toán 12

a) Đặt \(\left\{ \begin{array}{l}
u = x + \ln x\\
dv = {x^2}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \left( {1 + \frac{1}{x}} \right)dv\\
v = \frac{1}{3}{x^3}
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
I = \frac{1}{3}{x^3}\left( {x + \ln x} \right) - \frac{1}{3}\smallint {x^3}\left( {1 + \frac{1}{x}} \right)dx\\
 = \frac{1}{3}{x^4} + \frac{1}{3}{x^3}\ln x - \frac{1}{3}\smallint \left( {{x^3} + {x^2}} \right)dx\\
 = \frac{1}{3}{x^4} + \frac{1}{3}{x^3}\ln x - \frac{1}{3}\left( {\frac{1}{4}{x^4} + \frac{1}{3}{x^3}} \right) + C4\\
 = \frac{1}{4}{x^4} - \frac{1}{9}{x^3} + \frac{1}{3}{x^3}\ln x + C
\end{array}\)

b) \(J = \int {(x + {{\sin }^2}x)\sin xdx} \)

Đặt \(\left\{ \begin{array}{l}
u = x + \sin 2x\\
dv = \sin xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = 1 + 2\sin x\cos x\\
v =  - \cos x
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
J =  - \cos x\left( {x + {{\sin }^2}x} \right) + \smallint \cos x\left( {1 + 2\sin x\cos x} \right)dx\\
 =  - x\cos x - \cos x{\sin ^2}x + \smallint \cos xdx + 2\smallint {\cos ^2}x\sin xdx\\
 =  - x\cos x - \cos x{\sin ^2}x + \sin x - 2\smallint {\cos ^2}xd\left( {\cos x} \right)\\
 =  - x\cos x - \cos x{\sin ^2}x + \sin x - \frac{2}{3}{\cos ^3}x + C
\end{array}\)

c) \(K = \mathop \smallint \nolimits (x + {e^x}){e^{2x}}dx\)

Đặt \(\left\{ \begin{array}{l}
u = x + ex\\
dv = {e^{2x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \left( {1 + {e^x}} \right)dv\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
K = \frac{1}{2}\left( {x + {e^x}} \right){e^{2x}} - \frac{1}{2}\smallint {e^{2x}}\left( {1 + {e^x}} \right)dx\\
 = \frac{1}{2}\left( {x + {e^x}} \right){e^{2x}} - \frac{1}{2}\smallint \left( {{e^{2x}} + {e^{3x}}} \right)dx\\
 = \frac{1}{2}x{e^{2x}} + \frac{1}{2}{e^{3x}} - \frac{1}{4}{e^{2x}} - \frac{1}{6}{e^{3x}} + C\\
 = \frac{1}{2}x{e^{2x}} + \frac{1}{3}{e^{3x}} - \frac{1}{4}{e^{2x}} + C
\end{array}\)

d) \(F = \int {(x + \sin x)\frac{{dx}}{{{{\cos }^2}x}}} \)

Đặt \(\left\{ \begin{array}{l}
u = x + \sin x\\
dv = \frac{{dx}}{{{{\cos }^2}x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \left( {1 + \cos x} \right)dv\\
v = \tan x
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
F = \left( {x + \sin x} \right)\tan x - \int {\left( {1 + \cos x} \right)\tan xdx} \\
 = \left( {x + \sin x} \right)\tan x - \int {\left( {\frac{{\sin x}}{{\cos x}} + \sin x} \right)dx} \\
 = \left( {x + \sin x} \right)\tan x + \ln \left| {\cos x} \right| + \cos x + C
\end{array}\)

-- Mod Toán 12 HỌC247