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Question Number 145324 by mathdanisur last updated on 04/Jul/21

$$\int \frac{dx}{1+x^6}=?$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$\frac{1}{6}\mathrm{atan}\left(\frac{x}{1-x^2}\right)+\frac{1}{3}\mathrm{atan}x+\frac{\sqrt{3}}{12}\ln \mid \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}\mid +\mathrm{C}$$

Commented by mathdanisur last updated on 04/Jul/21

$$ThanksSer,solutionpossible$$

Answered by Canebulok last updated on 04/Jul/21

$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$let:$$$$\Rightarrow P^{}=x^3\Rrightarrow {}^3\sqrt{P^2}=x^2$$$$\Rightarrow dP=3\left({}^3\sqrt{P^2}\right)dx$$$$\because$$$$\Rightarrow \frac{1}{3}\cdot \int \frac{dP}{(1+P^2)\left({}^3\sqrt{P^2}\right)}=\phi$$$$$$$$Bysubstitution,$$$$let:$$$$\Rightarrow u=P^2\Rrightarrow \sqrt{u}=P$$$$\Rightarrow du=2PdP\Rrightarrow \frac{du}{2\sqrt{u}}=dP$$$$\because$$$$\Rightarrow \frac{1}{6}\cdot \int \frac{du}{(1+u)\left({}^6\sqrt{u^5}\right)}=\phi$$$$$$$$By‘‘I.B.P{.}^{″},$$$$let:$$$$\Rightarrow k=\frac{1}{{}^6\sqrt{u^5}}$$$$\Rightarrow dk=-\frac{5}{6\left({}^6\sqrt{u^{11}}\right)}$$$$\Rightarrow dz=\frac{du}{(u+1)}$$$$\Rightarrow z=ln(u+1)$$$$\because$$$$\Rightarrow \phi =\frac{ln(u+1)}{{}^6\sqrt{u^5}}+\left(\frac{5}{6}\right)\cdot \int \frac{ln(u+1)}{{}^6\sqrt{u^{11}}}du$$$$\Rightarrow \phi ={2\tan }^{-1}\left({}^6\sqrt{u}\right)-2{\left({}^6\sqrt{-1}\right)}^5\cdot {\tanh }^{-1}\left({}^6\sqrt{-u}\right)-2\left({}^6\sqrt{-1}\right)\cdot {\tanh }^{-1}({(-1)}^{5/6}\cdot \left({}^6\sqrt{u}\right)))+C$$$$Bysubstitutingbackagain,$$$$\Rightarrow \phi ={2\tan }^{-1}\left({}^3\sqrt{P}\right)-2{\left({}^6\sqrt{-1}\right)}^5\cdot {\tanh }^{-1}\left({}^6\sqrt{-1}{\cdot }^3\sqrt{P}\right)-2\left({}^6\sqrt{-1}\right)\cdot {\tanh }^{-1}({(-1)}^{5/6}\cdot \left({}^3\sqrt{P}\right))+C$$$$Thus;$$$$\Rightarrow \phi ={2\tan }^{-1}\left(x\right)-2{\left({}^6\sqrt{-1}\right)}^5\cdot {\tanh }^{-1}({}^6\sqrt{-1}\cdot x)-2\left({}^6\sqrt{-1}\right)\cdot {\tanh }^{-1}({(-1)}^{5/6}\cdot \left(x\right))+C$$$$$$

Commented by mathdanisur last updated on 04/Jul/21

$$ThanksSer,coolalot$$

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