Question Number 7819 by akmishra last updated on 17/Sep/16
$${let}\mathrm{2}{x}+\mathrm{1}={t} \\ $$$$\:\:\:\:{x}=\left({t}−\mathrm{1}\right)/\mathrm{2} \\ $$$${dx}={dt}/\mathrm{2} \\ $$$$\int\left({t}−\mathrm{1}\right)\sqrt{{t}}/\mathrm{2}{dt}/\mathrm{2} \\ $$$$\mathrm{1}/\mathrm{4}\int\left({t}\sqrt{{t}}−\sqrt{{t}}\right){dt} \\ $$$${continue}…..{it}.. \\ $$$$\:\: \\ $$$$ \\ $$$$ \\ $$
Answered by prakash jain last updated on 02/Oct/16
$$\mathrm{1}/\mathrm{4}\int\left({t}\sqrt{{t}}−\sqrt{{t}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({t}^{\mathrm{3}/\mathrm{2}} −{t}^{\mathrm{1}/\mathrm{2}} \right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{t}^{\mathrm{5}/\mathrm{2}} }{\mathrm{5}/\mathrm{2}}−\frac{{t}^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}/\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{10}}{t}^{\mathrm{5}/\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}/\mathrm{2}} +{C} \\ $$