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Question-141530




Question Number 141530 by sarkor last updated on 20/May/21
Commented by mohammad17 last updated on 20/May/21
=∫ ((x^2 −9+9)/( (√(x−3))))dx=∫ (((x−3)(x+3))/( (√(x−3))))dx+∫ (9/( (√(x−3))))dx    =∫((√(x−3)))(x+3)dx+∫9(x−3)^(−(1/2)) dx    =∫((√(x−3)))(x−3+6)dx+9∫(x−3)^(−(1/2)) dx    =∫(x−3)^(3/2) dx+6∫ (x−3)^(1/2) dx+9∫ (x−3)^(−(1/2)) dx    =(2/5)(√((x−3)^5 ))+4(√((x−3)^3 ))+18(√((x−3)))+C    by:: ⟨m.o⟩
$$=\int\:\frac{{x}^{\mathrm{2}} −\mathrm{9}+\mathrm{9}}{\:\sqrt{{x}−\mathrm{3}}}{dx}=\int\:\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)}{\:\sqrt{{x}−\mathrm{3}}}{dx}+\int\:\frac{\mathrm{9}}{\:\sqrt{{x}−\mathrm{3}}}{dx} \\ $$$$ \\ $$$$=\int\left(\sqrt{{x}−\mathrm{3}}\right)\left({x}+\mathrm{3}\right){dx}+\int\mathrm{9}\left({x}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$ \\ $$$$=\int\left(\sqrt{{x}−\mathrm{3}}\right)\left({x}−\mathrm{3}+\mathrm{6}\right){dx}+\mathrm{9}\int\left({x}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$ \\ $$$$=\int\left({x}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx}+\mathrm{6}\int\:\left({x}−\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}+\mathrm{9}\int\:\left({x}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$ \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{5}} }+\mathrm{4}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{3}} }+\mathrm{18}\sqrt{\left({x}−\mathrm{3}\right)}+{C} \\ $$$$ \\ $$$${by}::\:\langle{m}.{o}\rangle \\ $$

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