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Câu hỏi: Biết $\int{f\left( x \right)dx}=3x\cos \left( 2x-5 \right)+C$. Tìm khẳng định đúng trong các khẳng định sau
A. $\int{f\left( 3x \right)dx}=3x\cos \left( 6x-5 \right)+C$
B. $\int{f\left( 3x \right)dx}=9x\cos \left( 6x-5 \right)+C$
C. $\int{f\left( 3x \right)dx}=9x\cos \left( 2x-5 \right)+C$
D. $\int{f\left( 3x \right)dx}=3x\cos \left( 2x-5 \right)+C$
A. $\int{f\left( 3x \right)dx}=3x\cos \left( 6x-5 \right)+C$
B. $\int{f\left( 3x \right)dx}=9x\cos \left( 6x-5 \right)+C$
C. $\int{f\left( 3x \right)dx}=9x\cos \left( 2x-5 \right)+C$
D. $\int{f\left( 3x \right)dx}=3x\cos \left( 2x-5 \right)+C$
Đặt $t=3x\Rightarrow dt=3dx\Leftrightarrow dx=\dfrac{dt}{3}$
$I=\int{f\left( 3x \right)dx}=\dfrac{1}{3}\int{f\left( t \right)dt}=\dfrac{1}{3}\left[ 3t.\cos \left( 2t-5 \right)+{C}' \right]=t.\cos \left( 2t-5 \right)+\dfrac{{{C}'}}{3}$
$ =3x.\cos \left( 6x-5 \right)+C$
$I=\int{f\left( 3x \right)dx}=\dfrac{1}{3}\int{f\left( t \right)dt}=\dfrac{1}{3}\left[ 3t.\cos \left( 2t-5 \right)+{C}' \right]=t.\cos \left( 2t-5 \right)+\dfrac{{{C}'}}{3}$
$ =3x.\cos \left( 6x-5 \right)+C$
Đáp án A.