Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 133109 by EDWIN88 last updated on 18/Feb/21

$$\int \frac{dx}{(x^4+1)\sqrt[4]{x^4+2}}?$$

Commented by liberty last updated on 20/Feb/21

$$\mathrm{I}=\int \frac{\mathrm{dx}}{({\mathrm{x}}^4+1)\sqrt[4]{{\mathrm{x}}^4+2}}$$$$\mathrm{I}=\int \frac{\frac{1}{{\mathrm{x}}^5}\mathrm{dx}}{(1+\frac{1}{{\mathrm{x}}^4})\sqrt[4]{1+\frac{2}{{\mathrm{x}}^4}}}$$$$\mathrm{I}=-\frac{1}{4}\int \frac{\mathrm{d}\left(\frac{1}{{\mathrm{x}}^4}\right)}{(1+\frac{1}{{\mathrm{x}}^4})\sqrt[4]{1+\frac{2}{{\mathrm{x}}^4}}}$$$$\mathrm{let}\frac{1}{{\mathrm{x}}^4}=\mathrm{y}$$$$\mathrm{I}=-\frac{1}{4}\int \frac{\mathrm{dy}}{(1+\mathrm{y})\sqrt[4]{1+2\mathrm{y}}}$$$$\mathrm{let}\mathrm{again}\mathrm{z}=\sqrt[4]{1+2\mathrm{y}};\mathrm{y}=\frac{{\mathrm{z}}^4-1}{2}$$$$\mathrm{I}=-\frac{1}{4}\int \frac{{2\mathrm{z}}^3\mathrm{dz}}{(1+\frac{{\mathrm{z}}^4-1}{2})\mathrm{z}}$$$$\mathrm{I}=-\int \frac{{\mathrm{z}}^2}{{\mathrm{z}}^4+1}\mathrm{dz}=-\int \frac{1}{{\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2}}\mathrm{dz}$$$$\mathrm{I}=-\frac{1}{2}\int \frac{(1+\frac{1}{{\mathrm{z}}^2})+(1-\frac{1}{{\mathrm{z}}^2})}{{\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2}}\mathrm{dz}$$$$\mathrm{I}=-\frac{1}{2}[\int \frac{\mathrm{d}(\mathrm{z}-\frac{1}{\mathrm{z}})}{{(\mathrm{z}-\frac{1}{\mathrm{z}})}^2+2}+\int \frac{\mathrm{d}(\mathrm{z}+\frac{1}{\mathrm{z}})}{{(\mathrm{z}+\frac{1}{\mathrm{z}})}^2-2}]$$$$\mathrm{I}=-\frac{1}{2}[\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{z}-\frac{1}{\mathrm{z}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \mid \frac{\mathrm{z}+\frac{1}{\mathrm{z}}-\sqrt{2}}{\mathrm{z}+\frac{1}{\mathrm{z}}+\sqrt{2}}\mid +\mathrm{c}$$$$\mathrm{I}=-\frac{1}{2\sqrt{2}}\arctan \left(\frac{{\mathrm{z}}^2-1}{\mathrm{z}\sqrt{2}}\right)-\frac{1}{4\sqrt{2}}\ln \mid \frac{{\mathrm{z}}^2-\mathrm{z}\sqrt{2}+1}{{\mathrm{z}}^2+\mathrm{z}\sqrt{2}+1}\mid +\mathrm{c}$$$$\mathrm{I}=-\frac{1}{2\sqrt{2}}\arctan \left(\frac{\sqrt{1+2\mathrm{y}}-1}{\sqrt{2}\sqrt[4]{1+2\mathrm{y}}}\right)-\frac{1}{4\sqrt{2}}\ln \mid \frac{\sqrt{1+2\mathrm{y}}-\sqrt{2}\sqrt[4]{1+2\mathrm{y}}+1}{\sqrt{1+2\mathrm{y}}+\sqrt{2}\sqrt[4]{1+2\mathrm{y}}+1}\mid +\mathrm{c}$$$$\mathrm{I}=-\frac{1}{2\sqrt{2}}\arctan \left(\frac{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4+2}-1}{\mathrm{x}\sqrt{2}\sqrt[4]{{\mathrm{x}}^4+2}}\right)-\frac{1}{4\sqrt{2}}\ln \mid \frac{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4+2}-\mathrm{x}\sqrt{2}\sqrt{{\mathrm{x}}^4+2}+1}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4+2}+\mathrm{x}\sqrt{2}\sqrt[4]{{\mathrm{x}}^4+2}+1}\mid +\mathrm{c}$$$$$$

Answered by MJS_new last updated on 19/Feb/21

$$\int \frac{dx}{(x^4+1)\sqrt[4]{x^4+2}}=$$$$[t=\frac{\sqrt[4]{x^4+2}}{x}\iff x=\sqrt[4]{\frac{2}{t^4-1}}\to dx=-\frac{x^2\sqrt[4]{{(x^2+2)}^3}}{2}dt]$$$$=-\int \frac{t^2}{t^4+1}dt$$$$\mathrm{and}\mathrm{this}\mathrm{should}\mathrm{be}‘‘{\mathrm{easy}}^{″}$$

Answered by john_santu last updated on 19/Feb/21

Terms of Service

Privacy Policy

Contact: info@tinkutara.com