How do you integrate $arctan (\sqrt{x}) d x$ $arctan(sqrt(x))dx$ ?

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Answer 1 · 2016-07-21T23:12:45

$\int arctan (\sqrt{x}) d x = (x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $int arctan(sqrt(x))dx = (x+1)arctan(sqrt(x))-sqrt(x)+C$

Making $y = \sqrt{x}$ $y = sqrt(x)$ we have

$d y = \frac{1}{2} \frac{d x}{\sqrt{x}} = \frac{1}{2} \frac{d x}{y}$ $dy = 1/2(dx)/sqrt(x) = 1/2 (dx)/y$

$\int arctan (\sqrt{x}) d x \equiv \int 2 y arctan (y) d y$ $int arctan(sqrt(x))dx equiv int 2y arctan(y)dy$

Now

$\frac{d}{d y} (y^{2} arctan (y)) = 2 y arctan (y) + \frac{y^{2}}{1 + y^{2}}$ $d/(dy)(y^2 arctan(y))=2yarctan(y)+y^2/(1+y^2)$

but

$\frac{y^{2}}{1 + y^{2}} = 1 - \frac{1}{1 + y^{2}}$ $y^2/(1+y^2) = 1-1/(1+y^2)$

so

$\frac{d}{d y} (y^{2} arctan (y)) = 2 y arctan (y) + 1 - \frac{1}{1 + y^{2}}$ $d/(dy)(y^2 arctan(y))=2yarctan(y)+1-1/(1+y^2)$

Finally

$\int 2 y arctan (y) d y = y^{2} arctan (y) - y + arctan (y) + C$ $int 2y arctan(y)dy=y^2arctan(y)-y+arctan(y)+C$

or

$\int arctan (\sqrt{x}) d x = (x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $int arctan(sqrt(x))dx = (x+1)arctan(sqrt(x))-sqrt(x)+C$

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