Question Number 137695 by Mathspace last updated on 05/Apr/21
$${find}\:\:\int\:\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}{dx} \\ $$
Answered by bemath last updated on 05/Apr/21
$$\:\int\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\sqrt{{x}}−\sqrt{{x}+\mathrm{1}}\:\right)}{{x}−\left({x}+\mathrm{1}\right)}{dx} \\ $$$$=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}}\:\right){dx} \\ $$$$=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:{dx}−\int\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}}\:{dx} \\ $$$${I}_{\mathrm{1}} \:=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:{dx} \\ $$$${let}\:{x}+\mathrm{1}\:=\:{z}^{\mathrm{2}} \:\Rightarrow{dx}=\mathrm{2}{z}\:{dz} \\ $$$${I}_{\mathrm{1}} =\:\int\left(\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{2}{z}^{\mathrm{2}} \right){dz} \\ $$$${I}_{\mathrm{2}} =\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}}\:{dx} \\ $$$${let}\:{x}={q}^{\mathrm{2}} \:\Rightarrow{dx}=\mathrm{2}{q}\:{dq} \\ $$$${I}_{\mathrm{2}} =\:\int\:\left(\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{2}{q}^{\mathrm{2}} \right){dq} \\ $$