We are given the Indefinite integral ;
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{dx}{x-x^{3}}}[/tex]
Take x common from denominator ;
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{dx}{x(1-x^{2})}}[/tex]
Now , Put ;
[tex]{:\implies \quad \displaystyle \sf x^{2}=u}[/tex]
So that ;
[tex]{:\implies \quad \displaystyle \sf dx=\dfrac{du}{2\sqrt{u}}\quad and\quad x=\sqrt{u}}[/tex]
Now , putting the values ;
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{1}{\sqrt{u}(1-u)}\times \dfrac{1}{2}\times \dfrac{du}{\sqrt{u}}}[/tex]
Now , as constant can be taken out of the integrand, so now ;
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}\int \dfrac{du}{u(1-u)}}[/tex]
Using partial fraction decomposition, Rewrite the integral as ;
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}\int \left(\dfrac{1}{u}+\dfrac{1}{1-u}\right)du}[/tex]
As Integrals follow distributive property, so breaking the integral into two integrals, and continuing the integration
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}\left(\int \dfrac{1}{u}du+\int \dfrac{1}{1-u}du\right)}[/tex]
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}\bigg\{log|u|+\dfrac{log|1-u|}{(-1)}\bigg\}+C}[/tex]
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}(log|u|-log|1-u|)+C}[/tex]
[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{2}log\bigg|\dfrac{u}{1-u}\bigg| +C}[/tex]
Putting value of u ;
[tex]{:\implies \quad \therefore \quad \underline{\underline{\displaystyle \bf \int \dfrac{dx}{x-x^{3}}=\dfrac{1}{2}log\bigg|\dfrac{x^2}{1-x^{2}}\bigg| +C}}}[/tex]
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