For a continuous function f, the value ∫0∞ fxn+x−nlogxx+11+x2dx is
π2
0
-π
2π
Putting 1x=t⇒dx=−1t2dt, so
I=∫0∞ fxn+x−nlogxdxx=−∫∞0 ft−n+tnlog1ttdtt2=−∫0∞ ft−n+tnlogtdtt=−I⇒ 2I=0⇒I=0
The given integral is equal to
∫0∞ 11+x2dx=tan−1x0∞=π2