First slide
Evaluation of definite integrals
Question

For a continuous function f, the value 0fxn+xnlogxx+11+x2dx is

Moderate
Solution

Putting 1x=tdx=1t2dt, so

I=0fxn+xnlogxdxx=0ftn+tnlog1ttdtt2=0ftn+tnlogtdtt=I   2I=0I=0

The given integral is equal to 

011+x2dx=tan1x0=π2

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