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Câu hỏi: Tìm họ các nguyên hàm của hàm số $f\left( x \right)=\dfrac{\sin x-\cos x}{{{\left( \sin x+\cos x \right)}^{2}}-4}.$
A. $\dfrac{1}{4}\ln \left( \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x} \right)+C.$
B. $-\dfrac{1}{4}\ln \left( \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x} \right)+C.$
C. $\dfrac{1}{4}\ln \left( \dfrac{2+\sin x-\cos x}{2-\sin x+\cos x} \right)+C.$
D. $\dfrac{1}{4}\ln \left( \dfrac{\sin x+\cos x+2}{\sin x+\cos x-2} \right)+C.$
A. $\dfrac{1}{4}\ln \left( \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x} \right)+C.$
B. $-\dfrac{1}{4}\ln \left( \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x} \right)+C.$
C. $\dfrac{1}{4}\ln \left( \dfrac{2+\sin x-\cos x}{2-\sin x+\cos x} \right)+C.$
D. $\dfrac{1}{4}\ln \left( \dfrac{\sin x+\cos x+2}{\sin x+\cos x-2} \right)+C.$
$\int{f\left( x \right)dx=\int{\dfrac{\sin x-\cos x}{{{\left( \sin x+\cos x \right)}^{2}}-4}dx}}$
Đặt $\sin x+\cos x=t\Rightarrow (\sin x-\cos x)dx=-dt$
$\Rightarrow \int{f\left( x \right)dx=-\int{\dfrac{dt}{{{t}^{2}}-4}=}}\dfrac{1}{4}\int{(\dfrac{1}{t+2}-}\dfrac{1}{t-2})dt=\dfrac{1}{4}\ln \left| \dfrac{t+2}{t-2} \right|+C=\dfrac{1}{4}\ln \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x}+C$.
Đặt $\sin x+\cos x=t\Rightarrow (\sin x-\cos x)dx=-dt$
$\Rightarrow \int{f\left( x \right)dx=-\int{\dfrac{dt}{{{t}^{2}}-4}=}}\dfrac{1}{4}\int{(\dfrac{1}{t+2}-}\dfrac{1}{t-2})dt=\dfrac{1}{4}\ln \left| \dfrac{t+2}{t-2} \right|+C=\dfrac{1}{4}\ln \dfrac{2+\sin x+\cos x}{2-\sin x-\cos x}+C$.
Đáp án D.