Question Number 123145 by aurpeyz last updated on 23/Nov/20
$$\int\frac{{x}^{\mathrm{3}} }{\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)^{\mathrm{6}} }{dx} \\ $$
Commented by benjo_mathlover last updated on 23/Nov/20
$$\delta\left({x}\right)=\frac{\mathrm{1}}{\mathrm{12}}\int\:\frac{{d}\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)}{\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)^{\mathrm{6}} }\:=\:\frac{\mathrm{1}}{\mathrm{12}}\int{u}^{−\mathrm{6}} \:{du}\: \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{60}}\:{u}^{−\mathrm{5}} +\:{c}\:=\:−\frac{\mathrm{1}}{\mathrm{60}\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)^{\mathrm{5}} }\:+\:{c} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Nov/20
$$\int\frac{{x}^{\mathrm{3}} }{\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)^{\mathrm{6}} }{dx}=\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dt}}{{t}^{\mathrm{6}} }=−\frac{\mathrm{1}}{\mathrm{60}{t}^{\mathrm{5}} }=−\frac{\mathrm{1}}{\mathrm{60}\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{6}\right)^{\mathrm{5}} } \\ $$