Question Number 90138 by M±th+et£s last updated on 21/Apr/20
$$\int{x}\sqrt{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}}\:{dx} \\ $$
Commented by MJS last updated on 21/Apr/20
$$\mathrm{seems}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{solve} \\ $$
Commented by abdomathmax last updated on 21/Apr/20
$$\int\:{x}\sqrt{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}}{dx}=\int\:{x}\sqrt{\mathrm{3}}\sqrt{{x}^{\mathrm{3}} \:+\frac{\mathrm{7}}{\mathrm{4}}}{dc} \\ $$$${chngemrnt}\:\sqrt{{x}^{\mathrm{3}} +\frac{\mathrm{7}}{\mathrm{4}}}={x}+{t}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{7}}{\mathrm{4}}={x}^{\mathrm{2}} +\mathrm{2}{xt}\:+{t}^{\mathrm{2}} \:\Rightarrow{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{tx}\:−{t}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow \\ $$$${x}\:=\nu\left({t}\right)\:\Rightarrow\int\:{x}\sqrt{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}}{dx}\:=\sqrt{\mathrm{3}}\int{v}\left({t}\right){v}^{'} \left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{v}^{\mathrm{2}} \left({t}\right)+{c}\:\:\:{rest}\:{to}\:{finf}\:{x}\:{interms}\:{of}\:{t} \\ $$$${this}\:{work}\:{is}\:{eazy}\:{for}\:{sir}\:{mjs}... \\ $$