Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 95436 by 675480065 last updated on 25/May/20

$$\int \frac{{\mathrm{x}}^4\mathrm{dx}}{{\mathrm{x}}^8+{\mathrm{x}}^4+1}$$

Answered by Mr.D.N. last updated on 25/May/20

$$\mathrm{I}=\int \frac{{\mathrm{x}}^4\mathrm{dx}}{{\mathrm{x}}^8+{\mathrm{x}}^4+1}$$$$=\int \frac{{\mathrm{x}}^4\mathrm{dx}}{{\mathrm{x}}^4({\mathrm{x}}^4+1+\frac{1}{{\mathrm{x}}^4})}=\int \frac{1}{{\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}+1}\mathrm{dx}$$$$=\int \frac{1}{{({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})}^2-2+1}\mathrm{dx}=\int \frac{1}{{({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})}^2-{\left(1\right)}^2}\mathrm{dx}$$$$=\int \frac{1}{({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+1)({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}-1)}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}-1}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-2-1}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+2-1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-{\left(\sqrt{3}\right)}^2}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+{\left(1\right)}^2}\mathrm{dx}$$$$=\frac{1}{2}\left[\frac{1}{2\sqrt{3}}\log \frac{\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{3}}{\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{3}}\right]-\frac{1}{2}\left[\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{\mathrm{x}-\frac{1}{\mathrm{x}}}{\sqrt{3}}\right]+\mathrm{C}$$$$=\frac{1}{4\sqrt{3}}\log \frac{{\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1}{{\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1}-\frac{1}{2\sqrt{3}}{\tan }^{-1}\frac{{\mathrm{x}}^2-1}{\sqrt{3}\mathrm{x}}+\mathrm{C}//.$$$$$$

Commented by PRITHWISH SEN 2 last updated on 25/May/20

$$\mathrm{sir}\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}$$$$\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-{\left(\sqrt{3}\right)}^2}\mathrm{so}\mathrm{d}(\mathrm{x}+\frac{1}{\mathrm{x}})=1-\frac{1}{{\mathrm{x}}^2}$$$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-{\mathrm{a}}^2}\mathrm{cannot}\mathrm{be}\mathrm{applied}\mathrm{here}.$$

Commented by mathmax by abdo last updated on 25/May/20

$$\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3}=\int \frac{1-\frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3}\mathrm{dx}$$$$=\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3}\mathrm{dx}(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{u})+\int \frac{\mathrm{dx}}{{\mathrm{x}}^2({(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3)}$$$$\int \frac{\mathrm{du}}{{\mathrm{u}}^2-3}\left(\mathrm{solvable}\right)+\int \frac{\mathrm{dx}}{{\mathrm{x}}^2({\mathrm{x}}^2+2+\frac{1}{{\mathrm{x}}^2}-3)}$$$$=\int ...\mathrm{dx}+\int \frac{\mathrm{dx}}{{\mathrm{x}}^4+{2\mathrm{x}}^2+1-{3\mathrm{x}}^2}=\int ...\mathrm{dx}+\int \frac{\mathrm{dx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{and}$$$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}=\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2-{3\mathrm{x}}^2}=\int \frac{\mathrm{dx}}{({\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1)({\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1)}$$$$\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{solvable}\mathrm{by}\mathrm{decomposition}...$$$$$$

Commented by mathmax by abdo last updated on 25/May/20

$$\mathrm{sir}{\mathrm{m}}^{\mathrm{r}}\mathrm{dn}\mathrm{you}\mathrm{answer}\mathrm{is}\mathrm{not}\mathrm{correct}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com