Let I=∫−11log(x+x2+1)dx
Consider f(x)=log(x+x2+1)
Now, f(−x)=log(−x+1+x2)
⇒f(−x)=−log(x+1+x2)
⇒f(−x)=−f(x)
So, f(x) is an odd function.
∴I=0
JEE Main 2021 — Mathematics Calculus
The value of the integral ∫−11log(x+x2+1)dx is:
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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