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I was trying to solve the following question:

Evaluate: $$int{frac{1}{sqrt{x(1+x^2)}}dx}$$

This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.

But, I don't know how to start this. I want a hint to get started with this. Thanks!

Several CAS I used for $$I=int{frac{dx}{sqrt{x(1+x^2)}}}$$ express it as an awful elliptic integral but
it can be "simplified" as $$I=2 sqrt{x} ,, _2F_1left(frac{1}{4},frac{1}{2};frac{5}{4};-x^2right)+C$$ where appears the Gaussian or ordinary hypergeometric function.

If you want an acceptable approximation of it for the integral for $0 leq x leq 2$ , you could use the following Padé approximant built at $x=0$

$$2 sqrt x,, frac{1+frac{96645617 }{53568140}x^2+frac{31538874293
}{31069521200}x^4+frac{3758605721543 }{20195188780000}x^6+frac{791037744588979
}{123594555333600000}x^8} {1+frac{102002431 }{53568140}x^2+frac{108481355723
}{93208563600}x^4+frac{384016013641 }{1553476060000}x^6+frac{30608751719
}{2485561696000}x^8}$$ which is in error of $6.86times 10^{-6}text{ %}$ at $x=1$ but $1.13times 10^{-2}text{ %}$ at $x=2$.

For infinitely large values of the upper bound, we can expand the interand and integrate termwise to get
$$color{blue}{frac{Gamma left(frac{1}{4}right)^2}{2 sqrt{pi }}-sum_{n=0}^infty binom{-frac{1}{2}}{n}frac{ x^{-2 n-frac{1}{2}}}{2 n+frac{1}{2}}}$$ which gives the asymptotics and simple ways to approximate the value of the definite integral. For example, using $color{red}{10}$ terms only in the summation gives for fiteen significant figures
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
1.0 & color{blue}{1.85}008944498406 & 1.85407467730137 \
1.5 & color{blue}{2.136962}08793742 & 2.13696267480776 \
2.0 & color{blue}{2.32606419}317845 & 2.32606419421172 \
2.5 & color{blue}{2.46224022503}003 & 2.46224022503732 \
3.0 & color{blue}{2.565720296556}84 & 2.56572029655697 \
3.5 & color{blue}{2.64754771012530} & 2.64754771012530
end{array}
right)$$ and all of this being easy to compute since
$$a_n=binom{-frac{1}{2}}{n}frac{ x^{-2 n-frac{1}{2}}}{2 n+frac{1}{2}}implies a_{n+1}=-frac{(2 n+1) (4 n+1)}{2 (n+1) (4 n+5) x^2}, a_n$$

We can also express the solution in terms of the Beta Function and the Incomplete Beta Function as I cover here:

begin{align}
I&= int frac{1}{sqrt{xleft(1 + x^2right)}}:dx = int_0^x frac{t^{-frac{1}{2}}}{left(t^2 + 1 right)^{frac{1}{2}}} :dt\
&=frac{1}{2} left[ Bleft(frac{1}{2} - frac{-frac{1}{2} + 1}{2} , frac{-frac{1}{2} + 1}{2} right) - Bleft(frac{1}{1 + x^2};frac{1}{2} - frac{-frac{1}{2} + 1}{2} , frac{-frac{1}{2} + 1}{2} right) right] \
&=frac{1}{2} left[ Bleft( frac{1}{4} , frac{1}{4} right) - Bleft(frac{1}{1 + x^2};frac{1}{4} , frac{1}{4} right) right]
end{align}

Using the relationship between the Beta function and the Gamma Function we find that:

begin{equation}
Bleft( frac{1}{4} , frac{1}{4} right) = frac{Gammaleft(frac{1}{4}right)Gammaleft(frac{1}{4}right)}{Gammaleft(frac{1}{4} + frac{1}{4}right)} = frac{Gammaleft(frac{1}{4}right)^2}{Gammaleft(frac{1}{2}right)} = frac{Gammaleft(frac{1}{4}right)^2}{sqrt{pi}}
end{equation}

Thus our integral $I$ becomes:

begin{equation}
I = int frac{1}{sqrt{xleft(1 + x^2right)}}:dx = int_0^x frac{t^{-frac{1}{2}}}{left(t^2 + 1 right)^{frac{1}{2}}} :dt = frac{1}{2} left[ frac{Gammaleft(frac{1}{4}right)^2}{sqrt{pi}} - Bleft(frac{1}{1 + x^2};frac{1}{4} , frac{1}{4} right) right]
end{equation}