How do you integrate #int 1/(1+sqrtx) dx#?

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Answer 1 · 2016-10-14T04:27:23

Try something and see what sticks. I got

#2sqrtx - 2ln|1 + sqrtx| + C#

Let:

#u = sqrtx#
#du = 1/(2sqrtx)dx#
#dx = 2sqrtxdu = 2udu#

Therefore we now have:

#= int 1/(1 + u)*2udu#

#= 2int u/(1+u)du#

#= 2int (u + 1 - 1)/(1 + u)du#

#= 2int cancel((1 + u)/(1 + u))du - 2int 1/(1+u)du#

This is easy now. We'll get:

#= 2u - 2ln|1+u|#

Finally, just plug in #u = sqrtx# to get:

#= color(blue)(2sqrtx - 2ln|1 + sqrtx| + C)#