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