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Question Number 143210 by ArielVyny last updated on 11/Jun/21
∫017x+1+3x+1x+1dx
Answered by MJS_new last updated on 11/Jun/21
∫7x+1++3x+1x+1dx=∫7x+1x+1dx+∫3x+1x+1dx∫bx+1x+1dx=[t=(x+1)lnb→dx=dtlnb]=∫ettdt=Eit=Ei((x+1)lnb)+C⇒answeris≈17.0130
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