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Question Number 137695 by Mathspace last updated on 05/Apr/21

find ∫ (((2x+1)^3 )/( (√x)+(√(x+1))))dx

$${find}\:\:\int\:\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}{dx} \\ $$

Answered by bemath last updated on 05/Apr/21

∫ (((2x+1)^3 ((√x)−(√(x+1)) ))/(x−(x+1)))dx = ∫ (2x+1)^3 ((√(x+1)) −(√x) )dx = ∫ (2x+1)^3 (√(x+1)) dx−∫(2x+1)^3 (√x) dx I_1 = ∫ (2x+1)^3 (√(x+1)) dx let x+1 = z^2 ⇒dx=2z dz I_1 = ∫(2z^2 −1)^3 (2z^2 )dz I_2 = ∫ (2x+1)^3 (√x) dx let x=q^2 ⇒dx=2q dq I_2 = ∫ (2q^2 +1)^3 (2q^2 )dq

$$\:\int\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\sqrt{{x}}−\sqrt{{x}+\mathrm{1}}\:\right)}{{x}−\left({x}+\mathrm{1}\right)}{dx} \\ $$$$=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}}\:\right){dx} \\ $$$$=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:{dx}−\int\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}}\:{dx} \\ $$$${I}_{\mathrm{1}} \:=\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:{dx} \\ $$$${let}\:{x}+\mathrm{1}\:=\:{z}^{\mathrm{2}} \:\Rightarrow{dx}=\mathrm{2}{z}\:{dz} \\ $$$${I}_{\mathrm{1}} =\:\int\left(\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{2}{z}^{\mathrm{2}} \right){dz} \\ $$$${I}_{\mathrm{2}} =\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{{x}}\:{dx} \\ $$$${let}\:{x}={q}^{\mathrm{2}} \:\Rightarrow{dx}=\mathrm{2}{q}\:{dq} \\ $$$${I}_{\mathrm{2}} =\:\int\:\left(\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{2}{q}^{\mathrm{2}} \right){dq} \\ $$

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