Question Number 141526 by sarkor last updated on 20/May/21
Answered by rs4089 last updated on 20/May/21
$$\int\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}\:}\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}+\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\:{dx} \\ $$$${let}\:{x}^{\mathrm{2}} +\mathrm{2}{x}={t}\:\Rightarrow\left(\mathrm{2}{x}+\mathrm{2}\right){dx}={dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{t}}\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}\frac{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\mathrm{3}} }+{C} \\ $$