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Question

Evaluate the integral 20dxx+4x2 using substitution.

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Solution

20dxx+4x2=20dx(x2x4)
=20dx(x2x+14144)
=20dx[(x12)2174]
=20dx(172)2(x12)2
Let x12=tdx=dt
When x=0,t=12 and when x=2,t=32
20dx(172)2(x12)2=3212dt(172)2t2
=⎢ ⎢12(172)log172+t172t⎥ ⎥3212
=117log172+3217232log17212172+12
=117[log17+3173log17117+1]
=117[log17+3173×17+1171]
=117log[17+3+41717+3417]
=117log[20+41720417]
=117log(5+17517)
=117log[(5+17)(5+17)2517]
=117log[25+17+10178]
=117log(42+10178)
=117log(21+5174)

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