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Question Number 53381 by gunawan last updated on 21/Jan/19

∫ (1/([(x−1)^3 (x+2)^5 ]^(1/4) )) dx =

$$\int\:\frac{\mathrm{1}}{\left[\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} \right]^{\mathrm{1}/\mathrm{4}} }\:{dx}\:= \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

∫(dx/((x−1)^(3/4) (x+2)^(5/4) )) ∫(dx/((((x−1)/(x+2)))^(3/4) ×(x+2)^((5/4)+(3/4)) )) ∫(dx/((((x−1)/(x+2)))^(3/4) ×(x+2)^2 )) y=((x−1)/(x+2)) dy=(((x+2)(1)−(x−1)1)/((x+2)^2 ))dx dy=((x+2−x+1)/((x+2)^2 ))dx dy=((3dx)/((x+2)^2 )) ∫(dy/(3y^(3/4) ))=(1/3)∫(dy/y^(3/4) )=(1/3)×(y^(((−3)/4)+1) /(−(3/4)+1))+c =(4/3){((x−1)/(x+2))}^(1/4) +c

$$\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \left({x}+\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} } \\ $$$$\int\frac{{dx}}{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} ×\left({x}+\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$$$\int\frac{{dx}}{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} ×\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${y}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}} \\ $$$${dy}=\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{1}\right)−\left({x}−\mathrm{1}\right)\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{dx} \\ $$$$ \\ $$$${dy}=\frac{{x}+\mathrm{2}−{x}+\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{dx} \\ $$$${dy}=\frac{\mathrm{3}{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\int\frac{{dy}}{\mathrm{3}{y}^{\frac{\mathrm{3}}{\mathrm{4}}} }=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dy}}{{y}^{\frac{\mathrm{3}}{\mathrm{4}}} }=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{y}^{\frac{−\mathrm{3}}{\mathrm{4}}+\mathrm{1}} }{−\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{1}}+{c} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} +{c} \\ $$$$ \\ $$$$ \\ $$

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