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Evaluate `intsqrt(1+y^(2)).2ydy`

Answer» `"Let"I=intsqrt(1+y^(2)).2ydy`
`"Let" u =1+y^(2)` , then du =2ydy.
`I=int u^(1//2) du`
`=(u^((1//2)+1))/((1//2)+1)` [Integrate , using rule no. 3 with n=1/2]
=`(2)/(3)u^(3//2) +C` [Simpler form]
`=(2)/(3)(1+y^(2))^(3//2)+C` [Replace u by `1+y^(2)`]


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