Evaluate the following integrals:
x = t2
d(x) = 2t.dt
dx = 2t.dt
Substituting t and dt we get
⇒ 
⇒ 
⇒ 
⇒ 2(ln|1 + t|)
But t = √x
⇒ 2( ln|1 + √x|) + c.
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