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Question

Evaluate the integrals using substitution.
20xx+2 (Put x+2=t2).

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Solution

20xx+2dx
Let x+2=t2dx=2tdt
=22(t22).t.2t.dt=222(t42t2)dt=2[t552t33]22
=2[(3252×83)(4252.223)]=2(1615+8215)=32+16215

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