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[tex]

\int arctan \sqrt{x} dx

[/tex]

Since there is no elementary formula for integration of an inverse trigo function, we cannot manipulate the integrand in such a way as to integrate easily with one step of Integration by parts.(please verify)

So first, If we let [tex]u = arctan \sqrt{x}[/tex] and

[tex]dv = dx[/tex]

then

[tex]du = \frac{dx}{2 \sqrt{x} (x+1)}[/tex] and

[tex]v=x[/tex]

then we have

[tex]

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}

[/tex]

Now, Case 1, without simplifying [tex]\frac{x}{\sqrt {x}}[/tex]

let [tex]db = \frac{1}{2 \sqrt{x} (x+1)}dx[/tex]

and [tex]a = x[/tex].

Then [tex] da = dx[/tex]

and [tex] b = arctan \sqrt{x} [/tex]

so

[tex]

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - xarctan \sqrt{x} + \int arctan \sqrt{x} dx [/tex]

which of course will lead us nowhere but 0=0

Now, Case 2 - If we simplify [tex]\frac{x}{\sqrt {x}} = \sqrt{x}[/tex]

let [tex]a= \sqrt{x}[/tex] and [tex] db = \frac{dx}{1+x} [/tex],

then [tex]da = \frac{dx}{2 \sqrt{x}}[/tex] and [tex] b = ln (x+1)[/tex]

so

[tex]

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \frac{1}{2} ( \sqrt{x}ln (x+1) - \int \frac{ln |1+x| dx}{2 \sqrt{x}}) [/tex]

and if we do Integration by parts again, we will just get an infinite sequence of [tex] \sqrt{x}ln (x+1) [/tex] (please verify)

I'm really open to suggestions ^_^`