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Question Number 93144 by i jagooll last updated on 11/May/20

∫ ((x^3 −1)/(x^3 +6x^2 +10x)) dx

$$\int\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{6x}^{\mathrm{2}} +\mathrm{10x}}\:\mathrm{dx}\: \\ $$

Answered by MJS last updated on 11/May/20

((x^3 −1)/(x(x^2 +6x+10)))=1−(1/(10x))−((59x+94)/(10(x^2 +6x+10)))=  =1−(1/(10x))−((59(2x+6))/(20(x^2 +6x+10)))+((83)/(10(x^2 +6x+10)))  now it should be easy

$$\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{10}\right)}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}{x}}−\frac{\mathrm{59}{x}+\mathrm{94}}{\mathrm{10}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{10}\right)}= \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}{x}}−\frac{\mathrm{59}\left(\mathrm{2}{x}+\mathrm{6}\right)}{\mathrm{20}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{10}\right)}+\frac{\mathrm{83}}{\mathrm{10}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{10}\right)} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\ $$

Commented by i jagooll last updated on 11/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by MJS last updated on 11/May/20

you′re welcome

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

Commented by john santu last updated on 11/May/20

waw math lover sir

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