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Question Number 84126 by Roland Mbunwe last updated on 09/Mar/20

∫((5−x)/(1+(√((x−4)))))dx

$$\int\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{\left({x}−\mathrm{4}\right)}}\boldsymbol{{dx}} \\ $$

Answered by MJS last updated on 09/Mar/20

−∫((x−5)/(1+(√(x−4))))dx=∫(1−(√(x−4)))dx=  =x−(2/3)(x−4)^(3/2) +C

$$−\int\frac{{x}−\mathrm{5}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}=\int\left(\mathrm{1}−\sqrt{{x}−\mathrm{4}}\right){dx}= \\ $$$$={x}−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C} \\ $$

Commented by MJS last updated on 09/Mar/20

−((x−5)/(1+(√(x−4))))×((1−(√(x−4)))/(1−(√(x−4))))=−(((x−5)(1−(√(x−4))))/(5−x))=  =1−(√(x−4))

$$−\frac{{x}−\mathrm{5}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}×\frac{\mathrm{1}−\sqrt{{x}−\mathrm{4}}}{\mathrm{1}−\sqrt{{x}−\mathrm{4}}}=−\frac{\left({x}−\mathrm{5}\right)\left(\mathrm{1}−\sqrt{{x}−\mathrm{4}}\right)}{\mathrm{5}−{x}}= \\ $$$$=\mathrm{1}−\sqrt{{x}−\mathrm{4}} \\ $$

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