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Question Number 68094 by mhmd last updated on 04/Sep/19
∫dx/(π+e)x2x
Answered by MJS last updated on 05/Sep/19
∫dx(π+e)x2x=∫dx(π+e)x=−1(π+e)xln(π+e)+C[∫axdx=axlna;a=(π+e)−1]
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