Question Number 88206 by jagoll last updated on 09/Apr/20
$$\int\:\:\frac{\mathrm{x}+\mathrm{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\:\mathrm{dx}\: \\ $$
Answered by john santu last updated on 09/Apr/20
![= ∫ (x/(1+x^4 )) dx + ∫ (x^3 /(1+x^4 )) dx = (1/2)∫ ((2x)/(1+x^4 )) dx + (1/4)∫ ((4x^3 )/(1+x^4 )) dx = (1/2) ∫ ((d(x^2 ))/(1+(x^2 )^2 )) + (1/4)∫ ((d(1+x^4 ))/(1+x^4 )) = (1/4) ln (1+x^4 ) + (1/2)∫ (du/(1+u^2 )) , [ u =x^2 ] = (1/4)ln(1+x^4 ) +(1/2)arc tan (x^2 ) + c](https://www.tinkutara.com/question/Q88208.png)
$$=\:\int\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:+\:\int\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{{d}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:,\:\left[\:{u}\:={x}^{\mathrm{2}} \right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:+\frac{\mathrm{1}}{\mathrm{2}}{arc}\:\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)\:+\:{c} \\ $$
Commented by jagoll last updated on 09/Apr/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr} \\ $$