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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=2\text{x}\left( 3+{{e}^{x}} \right)$ là
A. $3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}-2{{\text{e}}^{x}}+C$
B. $6{{\text{x}}^{2}}+2\text{x}{{e}^{x}}+2{{\text{e}}^{x}}+C$
C. $3{{\text{x}}^{2}}+{{e}^{x}}-2\text{x}{{e}^{x}}+C$
D. $3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}+2{{\text{e}}^{x}}+C$
A. $3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}-2{{\text{e}}^{x}}+C$
B. $6{{\text{x}}^{2}}+2\text{x}{{e}^{x}}+2{{\text{e}}^{x}}+C$
C. $3{{\text{x}}^{2}}+{{e}^{x}}-2\text{x}{{e}^{x}}+C$
D. $3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}+2{{\text{e}}^{x}}+C$
Ta có $\int{f\left( x \right)d\text{x}}=\int{2\text{x}\left( 3+{{e}^{x}} \right)d\text{x}}=6\int{x\text{dx}}+2\int{x{{e}^{x}}d\text{x}}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{x}}d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Suy ra: $\int{f\left( x \right)d\text{x}}=3{{\text{x}}^{2}}+2\left( x{{e}^{x}}-\int{{{e}^{x}}d\text{x}} \right)=3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}-2{{\text{e}}^{x}}+C$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{x}}d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Suy ra: $\int{f\left( x \right)d\text{x}}=3{{\text{x}}^{2}}+2\left( x{{e}^{x}}-\int{{{e}^{x}}d\text{x}} \right)=3{{\text{x}}^{2}}+2\text{x}{{e}^{x}}-2{{\text{e}}^{x}}+C$.
Đáp án A.