Question-211879

Question Number 211879 by Spillover last updated on 23/Sep/24

Answered by BHOOPENDRA last updated on 23/Sep/24

= (((x−4))/(10(x^2 −4x+5))) +(1/(10))∫ ((x−4)/(x^3 −4x^2 +5x)) (by ostrogradsky′s method) =((x−4)/(10(x^2 −4x+5)))+(1/(10))[∫((4x−11)/(5(x^2 −4x+5)))dx−∫(4/(5x)) dx] =((x−4)/(10(x^2 −4x+5)))+((−4ln(∣x∣)+2ln(x^2 −4x+5)−3tan^(−1) (x−2))/(50))+c

$$=\:\frac{\left({x}−\mathrm{4}\right)}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}\:+\frac{\mathrm{1}}{\mathrm{10}}\int\:\frac{{x}−\mathrm{4}}{{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}}\:\left({by}\:{ostrogradsky}'{s}\:{method}\right) \\ $$$$\:\:=\frac{{x}−\mathrm{4}}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}+\frac{\mathrm{1}}{\mathrm{10}}\left[\int\frac{\mathrm{4}{x}−\mathrm{11}}{\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}{dx}−\int\frac{\mathrm{4}}{\mathrm{5}{x}}\:{dx}\right] \\ $$$$=\frac{{x}−\mathrm{4}}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}+\frac{−\mathrm{4}{ln}\left(\mid{x}\mid\right)+\mathrm{2}{ln}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)−\mathrm{3tan}^{−\mathrm{1}} \left({x}−\mathrm{2}\right)}{\mathrm{50}}+{c} \\ $$

Commented by Spillover last updated on 23/Sep/24