Bài tập 42 trang 175 SGK Toán 12 NC

a) Đặt \(u = \frac{1}{x} - 1 \Rightarrow du =  - \frac{1}{{{x^2}}}dx \)

\(\Rightarrow \frac{{dx}}{{{x^2}}} =  - du\)

Do đó: 

\(\begin{array}{l}
\int {\frac{1}{{{x^2}}}\cos \left( {\frac{1}{x} - 1} \right)dx}  =  - \int {\cos u} du\\
 =  - \sin u + C =  - \sin \left( {\frac{1}{x} - 1} \right) + C
\end{array}\)

b) Đặt \(u = 1 + {x^4} \Rightarrow du = 4{x^3}dx \)

\(\Rightarrow {x^3}dx = \frac{{du}}{4}\)

\(\begin{array}{l}
\int {{x^3}} {(1 + {x^4})^3}dx = \frac{1}{4}\int {{u^3}} du\\
 = \frac{{{u^4}}}{{16}} + C = \frac{1}{{16}}{(1 + {x^4})^4} + C
\end{array}\)

c) Đặt 

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{u = {x^3}}\\
{dv = {e^{2x}}dx}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = \frac{1}{3}dx}\\
{v = \frac{1}{2}{e^{2x}}}
\end{array}} \right.}\\
\begin{array}{l}
 \Rightarrow \int {\frac{{x{e^{2x}}}}{3}dx} \\
 = \frac{1}{6}x{e^{2x}} - \frac{1}{6}\int {{e^{2x}}dx} \\
 = \frac{1}{6}x{e^{2x}} - \frac{1}{{12}}{e^{2x}} + C
\end{array}
\end{array}\)

d) Đặt 

\(\begin{array}{l}
\left\{ \begin{array}{l}
u = {x^2}\\
dv = {e^x}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = 2xdx\\
v = {e^{x}}
\end{array} \right.\\
 \Rightarrow \int {{x^2}{e^x}dx}  = {x^2}{e^x} - 2\int {x{e^x}dx} \left( 1 \right)
\end{array}\)

Đặt 

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{u = x}\\
{dv = {e^x}dx}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = dx}\\
{v = {e^x}}
\end{array}} \right.}\\
\begin{array}{l}
 \Rightarrow \int {x{e^x}dx}  = x{e^x} - \int {{e^x}dx} \\
 = x{e^x} - {e^x} + C
\end{array}
\end{array}\)

Từ (1) suy ra: 

\(\begin{array}{l}
\int {{x^2}} {e^x}dx = {x^2}{e^x} - 2x{e^x} + 2{e^x} + C\\
 = {e^x}({x^2} - 2x + 2) + C
\end{array}\)

-- Mod Toán 12 HỌC247