Evaluate 
Let I = 
Let x2 + 1 = t …(i)
∴ d(x2 + 1) = dt
⇒ 2x dx = dt
⇒ x dx = dt/2
When x = 2; t = 22 + 1 = 5
When x = 3; t = 32 + 1 = 10
Substituting x2 + 1 and x dx in I
⇒ I = 
[
]
⇒ I = 
⇒ I = � log 2
= � log 2
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