√∫dx/[(x+1) √x]?

Answer 1

Let's try making the substitution
u = x^1/2
du = 1/2 * x^(-1/2) * dx
= 1/2 * dx / √x
So now we have to integrate
2 * du / (u^2 + 1)
which gives 2 * arctan(u)
= 2 * arctan(√x)
Your final answer is the sqrt of that.
ie √ [2 * arctan(√x) ]

Answer 2

Assume question reads:- I = ∫ dx / [ (x + 1).√x ]
Let x = t²
dx = 2t.dt
I = ∫ [ 2t / [ t.(t² + 1)] dt
I = 2.tan^(-1) t + C
I = 2.tan^(-1) (√x) + C

Answer 3

∫dx/[(x+1) √x] = 2*arctan(√x) + C

So... if you have that square root outside the integral as well, you would have:

√(2*arctan(√x) + C)
_______________

Answer 4

The answer is √(2arctan(√x) + c ).

You could obtain this by substituting x=t^2 in the given integral.