# Another integral I need help with $\frac{1+\sqrt x}{\sqrt x\sqrt{1-x^2}}$

Can someone help me figure out how to compute the integral $$\int \frac{1+\sqrt{x} } {\sqrt{x}\sqrt{1-x^2} }dx$$ Thanks in advance!

My attempts: I tried substituting $\sqrt{x} = u$ , but it gives me nothing. I also tried splitting the integral into the sum of two integrals, which leaves me with the calculation of the integral of $\frac{1}{\sqrt{x}\sqrt{1-x^2} }$ , and I don't know how to calculate it.

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@did: I added my attempts to the first message. – theMissingIngredient Jan 7 '13 at 9:52
Nearly all the parts of a proof were there... :-) – Did Jan 7 '13 at 13:03

It good to know that the expression of the form:

$$x^m(a+bx^n)^pdx$$ where $m,n,p,a,b$ are constant is called a differential binomial.

Theorem: The integral

$$\int x^m(a+bx^n)^pdx$$

can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:

$1.$ $p$ is an integer( $p>0$ use the Newton's binomial theorem and when $p<0$ then $x=t^k$ which $\text{lcm}(n,m)$).

$2.$ $\dfrac{m+1}{n}$ is an integer.

$3.$ $\dfrac{m+1}{n}+p$ is an integer.

It is clear that we should focus on the term $\int\frac{dx}{\sqrt{x}\sqrt{1-x^2}}$. By a simple investigation, above fact tells us this integral cannot be expressed in terms of elemantry functions. Try it!

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I didn't know that +1 (the term "differential binomial")! – amWhy Feb 22 '13 at 2:05

Your attempts failed with good reason.

Assuming everything's happening over $\mathbb{R}$, for this integral to make sense we must have $-1 \le x \le 1$ (in fact $0 \le x \le 1$), so a trigonometric substitution is valid. Putting $x = \sin \theta$, then we get

\begin{align} \int \dfrac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\, dx &= \int \dfrac{1+\sqrt{\sin \theta}}{\sqrt{\sin \theta} \cos \theta} \cdot \cos \theta\, d\theta \\ &= \int \dfrac{1+\sqrt{\sin \theta}}{\sqrt{\sin \theta}}\, d\theta \\ &= \int (\sin \theta)^{-\frac{1}{2}}\, d\theta + \theta + k\end{align}

for some constant of integration $k$.

But $(\sin \theta)^{-\frac{1}{2}}$ has no antiderivative expressible in terms of elementary functions. So there's no way of expressing the indefinite integral in terms of elementary functions.

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Great result. I did the problem in another way, Clive. + – Babak S. Jan 7 '13 at 9:59

The second part is $$\int\frac{\mathrm dx}{\sqrt{1-x^2}}\stackrel{(x=\sin\theta)}{=}\int\mathrm d\theta=\theta+C.$$ The first part is $$\int\frac{\mathrm dx}{\sqrt{x}\sqrt{1-x^2}}\stackrel{(x=\sqrt{y})}{=}\int\tfrac12y^{-3/4}(1-y)^{-1/2}\mathrm dy=\tfrac12\mathrm{B}(y;\tfrac14,\tfrac12)+C,$$ where $\mathrm B$ denotes the incomplete Beta function. Hence, on the interval $[0,1]$, $$\int\frac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\mathrm dx=\tfrac12\mathrm{B}(x^2;\tfrac14,\tfrac12)+\arcsin(x)+C.$$ For example, $$\int_0^1\frac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\mathrm dx=\frac{\Gamma(\frac14)\Gamma(\frac12)}{2\Gamma(\frac34)}+\frac\pi2=\frac\pi2(\sqrt{2\pi}\Gamma(\tfrac14)^2+1)\approx1.7128.$$

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Glad we have someone like you here. Indeed the Gamma function is one the functions which pave our way. Thanks, did. – Babak S. Jan 7 '13 at 10:51

HINT: $$\int \frac{1+\sqrt{x} } {\sqrt{x}\sqrt{1-x^2} }dx$$ $\sqrt{x}=t$ then $\frac{dx}{\sqrt{x}}=2dt,x^2=t^4$ $$\int \frac{1+t } {\sqrt{1-t^4} }dt$$

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How exactly does this help? I can't calculate this integral.... – theMissingIngredient Jan 7 '13 at 9:57
this is an elliptic integral of first kind – Adi Dani Jan 7 '13 at 10:08
For my friend here, +1 – Babak S. Jan 14 '13 at 4:50