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Question Number 86288 by john santu last updated on 28/Mar/20

$$\int \frac{dx}{{(1+x^4)}^{1/4}}$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$x^2=tana$$$$2xdx={sec}^{2}ada$$$$\int \frac{{sec}^{2}ada}{2\sqrt{tana}\times {\left({sec}^{2}a\right)}^{\frac{1}{4}}}$$$$\frac{1}{2}\int \frac{1}{{cos}^{2}a\times \frac{\sqrt{sina}}{\sqrt{cosa}}\times \frac{1}{\sqrt{cosa}}}$$$$\frac{1}{2}\int \frac{cosa}{(1-{sin}^{2}a)\sqrt{sina}}da$$$$b^2=sina$$$$2bdb=cosada$$$$\frac{1}{2}\int \frac{2bdb}{(1-b^2)b}$$$$\int \frac{db}{1-b^2}=\frac{1}{2}ln\left(\frac{1+b}{1-b}\right)+c$$$$\frac{1}{2}ln\left(\frac{1+\sqrt{sina}}{1-\sqrt{sina}}\right)+c$$$$tana=x^2\to sina=\frac{x^2}{\sqrt{1+x^4}}$$$$\mathit{s}\mathit{o}\mathit{a}\mathit{n}\mathit{s}=\frac{1}{2}\mathit{l}\mathit{n}\left(\frac{1+\frac{\mathit{x}}{{(1+{\mathit{x}}^4)}^{\frac{1}{4}}}}{1-\frac{\mathit{x}}{{(1+{\mathit{x}}^4)}^{\frac{1}{4}}}}\right)+\mathit{c}$$$$=\frac{1}{2}\mathit{l}\mathit{n}\left(\frac{{(1+{\mathit{x}}^4)}^{\frac{1}{4}}+\mathit{x}}{{(1+{\mathit{x}}^4)}^{\frac{1}{4}}-\mathit{x}}\right)+\mathit{c}$$$$$$

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