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Question Number 118347 by bramlexs22 last updated on 17/Oct/20
∫dxx+x3
Answered by benjo_mathlover last updated on 17/Oct/20
settingx=r6∫6r5r3+r2dr=∫6r3r+1dr=6∫(r2−r+1)−1r+1dr=6(13r3−12r2+r−ln∣r+1∣)+c=2r3−3r2+6r−6ln∣r+1∣+c=2x−3x3+6x6−6ln∣x6+1∣+c
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