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Question Number 186051 by Tom last updated on 31/Jan/23
∫0−1∣5x−5−x∣dx
Answered by Frix last updated on 31/Jan/23
=∫0−1∣2sinh(xln5)∣dx==−2∫0−1sinh(xln5)dx==−2ln5[cosh(xln5)]−10=165ln5
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