Evaluate the following integrals:
Assume e3x + 1 = t
⇒ d(e3x + 1) = dt
⇒ 3e3x=dt
⇒ e3x = 
Substituting t and dt in the given equation we get
⇒ 
⇒ 
⇒ 
But t = e3x + 1
∴ The above equation becomes
⇒ 
.
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