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Question

Evaluate x3+4x27x+5x+2dx

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Solution

x3+4x27x+5x+2dx

u=x+2,du=dx.

x=u2
x2=(u2)2=u2+44u

x3=(u2)3=u386u2+12u

x3+4x27x+5=u386u2+12u+4u2+1616u7u+14+5

=u32u211u+27

=u32u211u+27udu

=(u22u11+27u)du

=u33u211u+27lnu+C

=(x+2)33(x+2)211(x+2)+27|lnx+2+C(Ans)

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