Integral of $\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$

Question

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!

Answer 1

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{x\left(1 + x^2\right)}}\:dx = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt\\ &=\frac{1}{2} \left[ B\left(\frac{1}{2} - \frac{-\frac{1}{2} + 1}{2} , \frac{-\frac{1}{2} + 1}{2} \right) - B\left(\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[ B\left( \frac{1}{4} , \frac{1}{4} \right) - B\left(\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} B\left( \frac{1}{4} , \frac{1}{4} \right) = \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4} + \frac{1}{4}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\Gamma\left(\frac{1}{2}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}} \end{equation}

Thus our integral $I$ becomes:

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

Answer 2

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 $$I=2 \sqrt{x} \,\, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-x^2\right)+C$$ 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 $6.86\times 10^{-6}\text{ %}$ at $x=1$ but $1.13\times 10^{-2}\text{ %}$ 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 $$a_n=\binom{-\frac{1}{2}}{n}\frac{ x^{-2 n-\frac{1}{2}}}{2 n+\frac{1}{2}}\implies a_{n+1}=-\frac{(2 n+1) (4 n+1)}{2 (n+1) (4 n+5) x^2}\, a_n$$

Answer 3

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

Well, there is no elementary primitive, because the roots of the cubic under the root are not repeated: $0,i,-i$. It requires an elliptic integral: $$\int\frac1{\sqrt{x(1+x^2)}}\,dx=F\left(\cos^{-1}\frac{1-y}{1+y},m=\frac12\right)+C$$ On my side I prefer using the Byrd and Friedman tables rather than a CAS to work out elliptic integrals (the above formula is 239.00). Often the result is cleaner and the parameter $m$ ends up in the classic range of $[0,1]$.