Evaluate the following integrals:
Here (4x + 2) can be written as 2(2x + 1).
Now assume, x2 + x + 1 = t
d(x2 + x + 1) = dt
(2x + 1)dx = dt
⇒ 
⇒ 
⇒ 
⇒ 
But t = x2 + x + 1
⇒ 
.
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