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Question

Evaluate the integrals using substitution.
201x+4x2dx.


Solution

201x+4x2dx=201x+4x2(12)2+(12)2dx=2014+14[(x2x+14)]dx=2014+14(x12)2dx=201(172)2(x12)2dx

=12.172[log∣ ∣172+x12172x+12∣ ∣]20(dxa2x2=12aloga+xax)=117[log17+2x1172x+1]20=117[log17+3173log17117+1]=117log17+3173+17117+1(logmlogn=logmn)

=117log(17+3)(17+1)(173)(171)=117log20+41720417=117log5+17517×5+175+17

To remove square root term from denominator, we multiply by its conjugate i.e.,5+17 in both numerator and denominator.

=117log(5+17)2(5)2(17)2[(ab)(a+b)=a2b2]=117log25+17+10172517117log42+10178=117log21+5174


Mathematics

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