Integration, where have i gone wrong?

1. Feb 6, 2009

Dell

$$\int$$arctg$$\sqrt{x}$$dx

using$$\int$$udv=uv-$$\int$$vdu

dv=dx
v=x

u=arctg$$\sqrt{x}$$
du=u'dx=[1/(1+x)]dx

$$\int$$arctg$$\sqrt{x}$$dx=xarctg$$\sqrt{x}$$ - $$\int$$x*[1/(1+x)]dx
=xarctg$$\sqrt{x}$$ - $$\int$$[x/(1+x)]dx
=xarctg$$\sqrt{x}$$ - $$\int$$1-[1/(1+x)]dx
=xarctg$$\sqrt{x}$$ - x+ln|x+1|+c
but this is wrong, where have i made the mistake?

2. Feb 6, 2009

MathematicalPhysicist

The derivative of arctan(sqrt(x)) is (1/(1+x))*1/2(sqrt(x)), you can still solve this question by looking at:
sqrt(x)/(1+x)dx=2/3(d(x^3/2))/(1+(x^3/2)^(2/3))
sqrt(x)dx=2/3d(x^3/2)
and the integral of dy/(1+y^2/3)
can be solved by writing y^1/3=sinh(t).

I am not sure this will work, but I think that there is no easy way here.

3. Feb 6, 2009

thanks