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Câu hỏi: Nguyên hàm của hàm số $y=\dfrac{\sin 2x}{3+2\cos x}$ bằng
A. $3\ln \left| 3+2\cos x \right|-\cos x+C$
B. $\dfrac{3}{2}\ln \left( 3+2\cos x \right)-\cos x+C$
C. $-\dfrac{3}{2}\ln \left( 3+2\cos x \right)+\cos x+C$
D. $3\ln (3+2\cos x)+\cos x+C$
A. $3\ln \left| 3+2\cos x \right|-\cos x+C$
B. $\dfrac{3}{2}\ln \left( 3+2\cos x \right)-\cos x+C$
C. $-\dfrac{3}{2}\ln \left( 3+2\cos x \right)+\cos x+C$
D. $3\ln (3+2\cos x)+\cos x+C$
$H=\int{\dfrac{\sin 2x}{3+2\cos x}}dx=\int{\dfrac{2\sin x\cos x}{2\cos x+3}}dx=-2\int{\dfrac{\cos x}{2\cos x+3}}d(cosx)$
$\xrightarrow{2\cos x+3=t}H=-2\int{\dfrac{\dfrac{t-3}{2}}{t}d\left( \dfrac{t-3}{2} \right)=-\dfrac{1}{2}\int{\dfrac{t-3}{t}dt=-\dfrac{1}{2}\int{\left( 1-\dfrac{3}{t} \right)dt=-\dfrac{1}{2}\left( t-3\ln \left| t \right| \right)}+C}}$
$=-\dfrac{1}{2}\left( 2\cos x+3-3\ln \left| 2\cos x+3 \right| \right)+C=-\cos x-\dfrac{3}{2}+\dfrac{3}{2}\ln (2\cos x+3)+C$
$\xrightarrow{2\cos x+3=t}H=-2\int{\dfrac{\dfrac{t-3}{2}}{t}d\left( \dfrac{t-3}{2} \right)=-\dfrac{1}{2}\int{\dfrac{t-3}{t}dt=-\dfrac{1}{2}\int{\left( 1-\dfrac{3}{t} \right)dt=-\dfrac{1}{2}\left( t-3\ln \left| t \right| \right)}+C}}$
$=-\dfrac{1}{2}\left( 2\cos x+3-3\ln \left| 2\cos x+3 \right| \right)+C=-\cos x-\dfrac{3}{2}+\dfrac{3}{2}\ln (2\cos x+3)+C$
Đáp án B.