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Question Number 87279 by Ar Brandon last updated on 03/Apr/20

$$\int \frac{x^2}{1+x^4}dx$$

Commented by abdomathmax last updated on 03/Apr/20

$$complexmethodletdecomposeF\left(x\right)=\frac{x^2}{x^4+1}$$$$x^4+1=0\Rightarrow x^4=-1=e^{i(2k+1)\pi }\Rightarrow x_k=e^{i\frac{(2k+1)\pi }{4}}$$$$andk\in \left[[0,3]\right]sotherootsare$$$$x_0=e^{\frac{i\pi }{4}},x_1=e^{\frac{i3\pi }{4}},x_2=e^{i\frac{5\pi }{4}},x_3=e^{i\frac{7\pi }{4}}and$$$$F\left(x\right)=\sum _{k=0}^3\frac{a_k}{x-x_k}$$$$a_k=\frac{x_k^2}{4x_k^3}=-\frac{1}{4}{x}_k^3\Rightarrow F\left(x\right)=-\frac{1}{4}\sum _{k=0}^3\frac{x_k^3}{x-x_k}\Rightarrow$$$$\int \frac{x^2}{1+x^4}dx=-\frac{1}{4}\sum _{k=0}^3{x}_k^{3}ln(x-x_k)+C$$

Commented by Ar Brandon last updated on 04/Apr/20

$$PleasemayIknowhowyouswitchedfrom$$$$a_k=\frac{x_k^2}{4x_k^3}to{a}_k=-\frac{1}{4}{x}_k^3?$$

Answered by TANMAY PANACEA. last updated on 03/Apr/20

$$\int \frac{dx}{x^2+\frac{1}{x^2}}$$$$\frac{1}{2}\int \frac{1-\frac{1}{x^2}+1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$$$\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}+\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}$$$$\frac{1}{2}\times \frac{1}{2\sqrt{2}}ln\left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right)+\frac{1}{2}\times \frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$$

Commented by Ar Brandon last updated on 03/Apr/20

$$Right$$

Commented by TANMAY PANACEA. last updated on 03/Apr/20

$$thankhousir$$

Commented by peter frank last updated on 03/Apr/20

$$thankyou$$

Answered by redmiiuser last updated on 03/Apr/20

$$x=\sqrt{tanz}$$$$dx=\frac{\sec {}^{2}z.dz}{2.\sqrt{\tan z}}$$$$\int \frac{\tan z.\sec {}^{2}z.dz}{2.\sqrt{\tan z}.\sec {}^{2}z}$$$$=\frac{1}{2}\int \sqrt{\tan z}.dz$$$$\sec z=u$$$$du=\sec z.\tan z.dz$$$$\frac{1}{2}\int \frac{\sqrt{\tan z}}{\sec z.\tan z}du$$$$=\frac{1}{2}\int \frac{du}{u.\sqrt{u^2-1}}$$$$=\frac{1}{2}\int \frac{{(-1)}^{\frac{-1}{2}}.{(1-u^2)}^{\frac{-1}{2}}}{u}du$$$$=\frac{{(-1)}^{\frac{-1}{2}}}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2n!.u^{2n}}{2^{2n}.{(n!)}^2.\left(2n\right)}$$$$=\frac{{(-1)}^{\frac{-1}{2}}}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2n!.{(1+x^4)}^n}{2^{2n}.{(n!)}^2.2n}$$$$$$$$$$

Commented by redmiiuser last updated on 03/Apr/20

$$plscheck!$$

Commented by redmiiuser last updated on 03/Apr/20

$$Mr.ArBrandonI$$$$requestyoutocheck$$$$mysolution.$$

Commented by MJS last updated on 03/Apr/20

$$\mathrm{Mr}.\mathrm{this}\mathrm{forum}\mathrm{had}\mathrm{been}\mathrm{a}\mathrm{polite}\mathrm{place}\mathrm{before}$$$$\mathrm{you}\mathrm{arrived}.\mathrm{nobody}\mathrm{will}\mathrm{answer}\mathrm{you}\mathrm{if}\mathrm{you}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{change}\mathrm{your}\mathrm{behaviour}.$$

Commented by mr W last updated on 03/Apr/20

$$100\mathrm{\%}agree!$$

Commented by mr W last updated on 03/Apr/20

$$iwouldsay:ifyouaresurethatyou$$$$cansolveaquestion,thensolveit!$$$$peoplewillseeitandgivecomment$$$$whentheycanandwhentheylike.$$$$toforceotherpeopletocheckitor$$$$togivecommentisnotpolite,not$$$$cool,evenboring!andifyoucannot$$$$solveaquestionoryouarenotsure$$$$withyoursolution,youcanalso$$$$leaveit.nobodywillblameyouifyou$$$${don}^{\prime }tgiveanswertoeveryquestion.$$

Commented by Ar Brandon last updated on 04/Apr/20

$${Don}^{\prime }tgetit.$$

Commented by ajfour last updated on 04/Apr/20

$$ithinkthesesuggestionsare$$$$meantformrredmuiser..$$

Commented by MJS last updated on 04/Apr/20

$$\mathrm{yes}$$

Commented by redmiiuser last updated on 04/Apr/20

$$okthankseveryone$$$$forsuchbeautiful$$$$comments.Iurgedonly$$$$tocheckandI{didn}^{\prime }t$$$$forcedtocomment.$$

Commented by redmiiuser last updated on 04/Apr/20

$$Andalsoonewho$$$$givesquestionmust$$$$checkeveryanswer.$$$$Itshis/herdutyI$$$$onlyaskedtocheck$$$$mysolutionandin$$$$manyquestionspeople$$$$asktochecktheanswer$$$$andthereisnopoint$$$$toinsultanyone.$$$$Anywaythanks$$$$mr.MJS,mr.Wtogive$$$$suchcomments.$$$$$$

Commented by redmiiuser last updated on 04/Apr/20

$$Ifformethisforum$$$$isnotgettingapoliteplace$$$$thenpleaseexcuse$$$$meitsmytendency$$$$togetexcited.Iam$$$$sorryHonourable$$$$teachers.$$

Commented by Ar Brandon last updated on 04/Apr/20

$$I{don}^{\prime }tgetthemethod.$$

Commented by redmiiuser last updated on 04/Apr/20

$$whymistercanyoutell$$$$methepartwhereyou$$$$facedproblem$$

Commented by ajfour last updated on 05/Apr/20

Commented by redmiiuser last updated on 05/Apr/20

$$hahaha...nicesketch$$

Commented by ajfour last updated on 05/Apr/20

thankx, was meant 4 u.

Commented by Ar Brandon last updated on 05/Apr/20

$$Noworries.I^{\prime }llreviewitsome$$$$othertime.$$$$Irealisedyouloveusingthesummationseries$$$$whilesolvingintegrals.Haha$$

Answered by MJS last updated on 04/Apr/20

$$\int \frac{x^2}{x^4+1}dx=$$$$=\frac{\sqrt{2}}{4}\int \frac{x}{x^2-\sqrt{2}x+1}dx-\frac{\sqrt{2}}{4}\int \frac{x}{x^2+\sqrt{2}x+1}dx=$$$$=\frac{\sqrt{2}}{8}\ln (x^2-\sqrt{2}x+1)+\frac{\sqrt{2}}{4}\arctan (\sqrt{2}x-1)-$$$$-\frac{\sqrt{2}}{8}\ln (x^2+\sqrt{2}x+1)+\frac{\sqrt{2}}{4}\arctan (\sqrt{2}x+1)=$$$$=\frac{\sqrt{2}}{8}(\ln \frac{x^2-\sqrt{2}x+1}{x^2-\sqrt{2}x+1}+2(\arctan (\sqrt{2}x-1)+\arctan (\sqrt{2}x+1)))+C$$

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