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Question Number 206541 by luciferit last updated on 18/Apr/24
Answered by lepuissantcedricjunior last updated on 18/Apr/24
∫3x+5x2−10x+51dx=kk=32∫2x−5+103+5x2−10x+51dxk=3x2−10x+51+252∫dx(x−5)2+26k=3x2−10x+51+25226∫dx1+(x−526)2x−526=sin(ha)=>dx=26cos(ha)dak=3x2−10x+51+252argsin(x−526)+c
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