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e2xcosxdx=15e2x(2cosx+sinx)+C\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C

question 135

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e2xcosxdx=15e2x(2cosx+sinx)+C\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C .


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