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Verified answer
∫(2x+4)√(2x²+8x+1) dxMisal u=2x²+8x+1
du/dx=4x+8
du/dx=2(2x+4)
du/2=2x+4 dx
Maka:
∫ √u du/2
=>½∫ √u du
=>½ (⅔ u√u)
=>⅓ (2x²+8x+1)√(2x²+8x+1) + C (C)
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