Q23 of 865 Page 20

Evaluate the following integrals:

x = t2


d(x) = 2t.dt


dx = 2t.dt


Substituting t and dt we get


⇒


⇒


Add and subtract 1 from numerator


⇒


⇒


⇒


⇒ 2(t – ln|1 + t|)


But t = √x


⇒ 2(√x – ln|1 + √x|) + c


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