Menu Close

3x-3-x-1-3-dx-




Question Number 167006 by cortano1 last updated on 04/Mar/22
       ∫ ((3x^3 )/((x−1)^3 )) dx=?
$$\:\:\:\:\:\:\:\int\:\frac{\mathrm{3x}^{\mathrm{3}} }{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dx}=? \\ $$
Answered by MJS_new last updated on 04/Mar/22
∫((3x^3 )/((x−1)^3 ))dx=∫((3(t+1)^3 )/t^3 )dt=  =∫(3+(9/t)+(9/t^2 )+(3/t^3 ))dt=  =3t+9ln t −(9/t)−(3/(2t^2 ))=  =((2(2t^3 −6t−1))/(2t^2 ))+9ln t =  =((3(2x^3 −6x^2 +3))/(2(x−1)^2 ))+9ln ∣x−1∣ +C
$$\int\frac{\mathrm{3}{x}^{\mathrm{3}} }{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{dx}=\int\frac{\mathrm{3}\left({t}+\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{3}} }{dt}= \\ $$$$=\int\left(\mathrm{3}+\frac{\mathrm{9}}{{t}}+\frac{\mathrm{9}}{{t}^{\mathrm{2}} }+\frac{\mathrm{3}}{{t}^{\mathrm{3}} }\right){dt}= \\ $$$$=\mathrm{3}{t}+\mathrm{9ln}\:{t}\:−\frac{\mathrm{9}}{{t}}−\frac{\mathrm{3}}{\mathrm{2}{t}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}{t}^{\mathrm{3}} −\mathrm{6}{t}−\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} }+\mathrm{9ln}\:{t}\:= \\ $$$$=\frac{\mathrm{3}\left(\mathrm{2}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{9ln}\:\mid{x}−\mathrm{1}\mid\:+{C} \\ $$
Commented by peter frank last updated on 04/Mar/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *