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tính nguyên hàm:
5sinx+2cosx/sinx-6cosx+1
[laTEX]I = \frac{32}{37}\int \frac{(cosx+6sinx)dx}{sinx-6cosx+1} - \frac{7}{37}\int \frac{(sinx-6cosx+1)dx}{sinx-6cosx+1} + \frac{7}{37}\int \frac{dx}{sinx-6cosx+1} \\ \\ I = \frac{32}{37}I_1 - \frac{7}{37}I_2 + \frac{7}{37}I_3 \\ \\ I_1 = ln|sinx-6cosx+1| \\ \\ I_2 = x \\ \\ I_3 : tan(\frac{x}{2}) = u \Rightarrow du = \frac{1}{2cos^2(\frac{x}{2})}dx = \frac{1}{2}(1+u^2)dx \\ \\ sinx - 6cosx +1 = \frac{2u}{1+u^2} - \frac{6(1-u^2)}{1+u^2} + \frac{1+u^2}{1+u^2} \\ \\ \Rightarrow \int \frac{du}{2(2u - 6 +6u^2 + 1+ u^2)}[/laTEX]
đến đây bạn tự làm nốt được rồi