Question Number 20255 by tammi last updated on 24/Aug/17
$$\int\frac{{x}^{\mathrm{3}} {dx}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$
Answered by Tinkutara last updated on 24/Aug/17
$${x}\:−\:\mathrm{1}\:=\:{t}^{\mathrm{2}} \\ $$$${dx}\:=\:\mathrm{2}{tdt} \\ $$$${x}^{\mathrm{3}} \:=\:\left({t}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\int\frac{{x}^{\mathrm{3}} {dx}}{\:\sqrt{{x}\:−\:\mathrm{1}}}\:=\:\int\frac{\left({t}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{3}} ×\mathrm{2}{tdt}}{{t}}\:=\:\mathrm{2}\int\left({t}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{3}} {dt} \\ $$$$=\:\mathrm{2}\int\left({t}^{\mathrm{6}} \:+\:\mathrm{3}{t}^{\mathrm{4}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{1}\right){dt} \\ $$$$=\:\mathrm{2}\left(\frac{{t}^{\mathrm{7}} }{\mathrm{7}}\:+\:\frac{\mathrm{3}{t}^{\mathrm{5}} }{\mathrm{5}}\:+\:{t}^{\mathrm{3}} \:+\:{t}\right)\:+\:{C} \\ $$$$=\:\mathrm{2}\left(\frac{\left({x}−\mathrm{1}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }{\mathrm{7}}\:+\:\frac{\mathrm{3}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}\:+\:\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:\sqrt{{x}−\mathrm{1}}\right)\:+\:{C} \\ $$
Commented by tammi last updated on 25/Aug/17
$$\left.{t}\left.{hanksss}:\right)\right) \\ $$