$\int_{0}^{1} \sqrt[]{1+t^4}dt$

Question

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}

Answer 1

\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}