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

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

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