How to integrate the following? $$\int_{3/2}^{8/3}\frac{\sqrt {x+\sqrt {x+\sqrt x}}}{\sqrt{x^{2}+\sqrt{x^{2}+\sqrt{x^{2}}}}}dx$$ I am trying this question by writing the denominator as $\sqrt{x^{2}+\sqrt{x^{2}+x}}$ because $\sqrt{x^{2}}=x$. Now I am facing problem to simplify further. Please help me out. Also, I think we can't substitute $x=(tan\theta)^{4}$, because then the denominator will become very messy. So, please help me out to integrate the above expression.
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2Do you have reason to believe this has a closed-form solution? Or are you asking for numerical integration? – DominikS Oct 31 '23 at 09:34
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Hi, welcome to Math SE. Wolfram Alpha doesn't find a closed form. – J.G. Oct 31 '23 at 09:41
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Note that the simpler integral (without the denominator) is already very tricky. – DominikS Oct 31 '23 at 09:50
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3Those limits of integration seem suspicious. There may be a way to do the integral without finding its antiderivative. – eyeballfrog Oct 31 '23 at 10:32
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On yesterday, we had a much simpler problem (have a look here) for which @Mariusz Iwaniuk gave the solution (which is a monster).
For your case, if you plot the integrand for the range of interest, you should notice that it is pretty close to a straight line.
So, write $$\frac{\sqrt {x+\sqrt {x+\sqrt x}}}{\sqrt{x^{2}+\sqrt{x^{2}+\sqrt{x^{2}}}}}=\sum_{n=0}^\infty a_n\, (x-2)^n$$ I chose $2$ because $\frac 12\left(\frac{3}{2}+\frac{8}{3}\right)=2+\frac 1{12}$.
The coefficients are really ugly but the result will be $$\sum_{n=0}^\infty a_n\, \frac{2^{n+2}\times 3^{-n}+(-1)^ n\, 3\times 2^{-n}}{6 (n+1)}$$
Computing the partial sums $$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 0 & 0.901131 \\ 1 & 0.885235 \\ 2 & 0.891661 \\ 3 & 0.891215 \\ 4 & 0.891337 \\ 5 & 0.891325 \\ 6 & 0.891327 \\ \end{array} \right)$$
Claude Leibovici
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