The value of ∫logex+x2+1x2+1dx, is
2logex+x2+1+C
logex+x2+12+C
logx+x2+1+C
none of these
We have.
I=∫logcx+x2+1x2+1dx∣⇒ l=∫logcx+x2+1dlogex+x2+1⇒ I=12logpx+x2+12+C