Question Number 152029 by mathdanisur last updated on 25/Aug/21
Commented by ghimisi last updated on 25/Aug/21
$$\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{2}} \:\:{or}\:\:\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 25/Aug/21
$$\mathrm{Sorry}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\:\mathrm{1}-\mathrm{4x}^{\mathrm{2}} \\ $$
Answered by ghimisi last updated on 26/Aug/21
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \right)\sqrt[{\mathrm{3}}]{\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}\right)^{'} \centerdot\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} = \\ $$$$−\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{3}}} \mid_{\mathrm{0}} ^{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by mathdanisur last updated on 26/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser},\:\mathrm{but}\:\mathrm{ans}:\:-\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$