How do you integrate #arctan(sqrt(x))dx#?

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Answer 1 · 2016-07-21T23:12:45

#int arctan(sqrt(x))dx = (x+1)arctan(sqrt(x))-sqrt(x)+C#

Making #y = sqrt(x)# we have

#dy = 1/2(dx)/sqrt(x) = 1/2 (dx)/y#

#int arctan(sqrt(x))dx equiv int 2y arctan(y)dy#

Now

#d/(dy)(y^2 arctan(y))=2yarctan(y)+y^2/(1+y^2)#

but

#y^2/(1+y^2) = 1-1/(1+y^2)#

so

#d/(dy)(y^2 arctan(y))=2yarctan(y)+1-1/(1+y^2)#

Finally

#int 2y arctan(y)dy=y^2arctan(y)-y+arctan(y)+C#

or

#int arctan(sqrt(x))dx = (x+1)arctan(sqrt(x))-sqrt(x)+C#