Q66 of 865 Page 20

Evaluate the following integrals:

Assume ex – 1 = t2


d(ex – 1) = d(t2)


ex.dx = 2t.dt


⇒


ex = t2 + 1


⇒


Substituting t and dt


⇒


⇒


⇒


⇒


Add and subtract 1 in numerator


⇒


⇒


⇒


⇒


⇒ 2t – 2tan - 1(t) + c


But t = (ex – 1)1/2


⇒ 2(ex – 1) 1/2– 2tan - 1(ex – 1) 1/2 + c


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