Evaluate the following integrals:
Assume x2 + sin2x + 2x =t
d(x2 + sin2x + 2x) =dt
(2x + 2cos2x + 2)dx = dt
2(x + cos2x + 1)dx = dt
(x + cos2x + 1)dx =
Put t and dt in given equation we get
⇒ 
= 
But t = x2 + sin2x + 2x
= 
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