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Question Number 211579 by MrGaster last updated on 13/Sep/24

$$$$$$\int \frac{1}{(1-{\mathit{x}}^4)\sqrt{1+{\mathit{x}}^2}}\mathit{d}\mathit{x}.$$$$$$

Answered by lepuissantcedricjunior last updated on 13/Sep/24

$$\int \frac{1}{(1-{\mathit{x}}^2)(1+{\mathit{x}}^2)\sqrt{1+{\mathit{x}}^2}}\mathit{d}\mathit{x}=\mathit{k}$$$$\mathit{x}=\mathit{t}\mathit{a}\mathit{n}\mathit{t}=>\mathit{d}\mathit{x}=(1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t})\mathit{d}\mathit{t}$$$$\mathit{k}=\int \frac{\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathit{d}\mathit{t}}{(1-{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t})}=\int \frac{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{t}}{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{t}-{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{t}}\mathit{d}\mathit{t}$$$$=\int \frac{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{t}}{\mathit{c}\mathit{o}\mathit{s}2\mathit{t}}\mathit{d}\mathit{t}=\int \frac{1+\mathit{c}\mathit{o}\mathit{s}2\mathit{t}}{2\left(\mathit{c}\mathit{o}\mathit{s}2\mathit{t}\right)}\mathit{d}\mathit{t}$$$$=\frac{1}{2}\mathit{t}+\int \frac{1}{2\mathit{c}\mathit{o}\mathit{s}2\mathit{t}}\mathit{d}\mathit{t}$$$$\mathit{o}=\mathit{t}\mathit{a}\mathit{n}\mathit{t}=>\mathit{d}\mathit{t}=\frac{\mathit{d}\mathit{o}}{1+{\mathit{o}}^2}$$$$\mathit{m}=\int \frac{\mathit{d}\mathit{t}}{2(2{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{t}-1)}=\int \frac{\mathit{d}\mathit{o}}{2(\frac{2}{1+{\mathit{o}}^2}-1)\times (1+{\mathit{o}}^2)}$$$$=\int \frac{\mathit{d}\mathit{o}}{(4-2-2{\mathit{o}}^2)}=\int \frac{\mathit{d}\mathit{o}}{2(1-{\mathit{o}}^2)}$$$$=\frac{1}{2}\int \frac{\mathit{d}\mathit{o}}{(1-\mathit{o})(1+\mathit{o})}=\frac{1}{4}\int (\frac{1}{1-\mathit{o}}+\frac{1}{1+\mathit{o}})\mathit{d}\mathit{o}$$$$=\frac{1}{4}\mathit{l}\mathit{n}\mid \frac{1+\mathit{o}}{1-\mathit{o}}\mid +\mathit{c}$$$$\mathit{k}=\frac{1}{2}\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}+\frac{1}{4}\mathit{l}\mathit{n}\mid \frac{1+\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}}{1-\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}}\mid +c$$$$\mathit{k}=\frac{1}{2}\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\right)+\frac{1}{4}\mathit{l}\mathit{n}\mid \frac{1+\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\right)}{1-\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\right)}\mid +\mathit{c}$$$$$$

Commented by Frix last updated on 13/Sep/24

$$3^{\mathrm{rd}}\mathrm{line}:$$$$x=\tan t\wedge dx=(1+{\tan }^{2}t)dt$$$$\mathrm{leads}\mathrm{to}$$$$k=\int \frac{{\cos }^{3}t}{{\cos }^{2}t-{\sin }^{2}t}dt$$$$\Rightarrow \mathrm{everything}\mathrm{else}\mathrm{is}\mathrm{wrong}.$$

Answered by Frix last updated on 21/Sep/24

$$\int \frac{dx}{(1-x^4)\sqrt{1+x^2}}\stackrel{[t=\frac{\sqrt{2}x}{\sqrt{x^2+1}}]}{=}$$$$=\frac{\sqrt{2}}{4}\int \frac{t^2-2}{t^2-1}dt=\frac{\sqrt{2}}{8}\int (\frac{1}{t+1}-\frac{1}{t-1}+2)dt=$$$$=\frac{\sqrt{2}}{8}(2t+\ln \frac{t+1}{t-1})=$$$$=\frac{x}{2\sqrt{x^2+1}}+\frac{\sqrt{2}}{8}\ln \mid \frac{\sqrt{2}x+\sqrt{x^2+1}}{\sqrt{2}x-\sqrt{x^3+1}}\mid +C$$

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