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I tried by using $t^2$ = $tan\theta$ and then by inserting $t^4$ by $tan^2\theta$ in $\int_{0}^{1} \sqrt[]{1+t^4}dt$, I get $dt$ = $\frac{sec^2\theta\times d\theta}{2\times \sqrt[]{tan\theta}}$ & $\sqrt[]{1+t^4}$ = $sec\theta$. Thus the integration becomes $\int_{0}^{\frac{\pi}{4}} \frac{sec^3 \theta \times d\theta}{2\times \sqrt{tan\theta}}$ What to do after this step? I have tried to do this which I have mistakenly written in the answer column. \begin{align} \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\int_{0}^{1}\frac{1+t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}\frac{t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}t\cdot\frac{t^3}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\left[\frac12t\sqrt{1+t^4}\right]_{0}^{1}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac121\sqrt{1+1^4}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\ \implies \frac32\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\frac{1}{2}\sqrt{1+1^4}+\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\ \implies \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\frac{1}{3}\sqrt{1+1^4}+\frac23\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}.\\ \end{align}

Saradamani
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    This is not an elementary integral. – Artem Apr 04 '18 at 14:37
  • Sorry it is a definite integral but I believe I have got the problem – Saradamani Apr 04 '18 at 14:38
  • There is not simple way to evaluate this integral exactly without invoking special functions. The value of this integral (numerically) is 1.09 – Artem Apr 04 '18 at 14:39
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    WA sasy the answer is given by $$\frac{1}{3} \sqrt{2} \left(1+(1+i) F\left(\left.\frac{1}{2} i \log \left((1+2 i)-2 \sqrt{-1+i}\right)\right|-1\right)\right)$$ – Dr. Sonnhard Graubner Apr 04 '18 at 14:46
  • The integral is not elementary, but can be expressed in terms of gamma function only. See https://math.stackexchange.com/questions/2311583 – pisco Apr 04 '18 at 14:50
  • @Dr.onnhard Graubner Can you please show how to arrive at this expression? – Saradamani Apr 04 '18 at 14:54
  • Oh no no I did not duplicate it meaning copied it. This is how I tried earlier but I was stuck with the denominator $\sqrt{1+t^4}$ – Saradamani Apr 04 '18 at 15:14

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\begin{align} \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\int_{0}^{1}\frac{1+t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}\frac{t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}t\cdot\frac{t^3}{\sqrt{1+t^4}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\left[\frac12t\sqrt{1+t^4}\right]_{0}^{1}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac121\sqrt{1+1^4}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\ \implies \frac32\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\frac{1}{2}\sqrt{1+1^4}+\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\ \implies \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t &=\frac{1}{3}\sqrt{1+1^4}+\frac23\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}.\\ \end{align}

Saradamani
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