Practice Problems: Integration by Parts (Solutions)
Practice Problems: Integration by Parts (Solutions) Written by Victoria Kala ... I will show you two solutions. Solution I: First do the substitution w= lnx. Then dw= 1 x dxand x= ew. Then Z 2 1 (lnx)2 x3 dx= Z ln2 0 w e2w dw= Z ln2 0 w2e 2wdw Tabular is easy on this guy: Z ln2 0 w2e 2wdw= w 2 2 e 2w w 2 e 2w 1 4 e 2w ln2ln2 0 = e w 2 w2 + w+ 1 ...
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