Question Number 200915 by Rupesh123 last updated on 26/Nov/23
Answered by Frix last updated on 26/Nov/23
![Assume ∫(((x^(1/3) +1)^(4/5) )/x^(8/5) )dx=((p(x^(1/3) +1)^(q/5) )/x^(r/5) ) f′(x)=(d/dx)[((p(x^(1/3) +1)^(q/5) )/x^(r/5) )]= ((p(x^(1/3) +1)^((q/5)−1) ((q−3r)x^(1/3) −3r))/x^((r/5)+1) ) (q/5)−1=(4/5) ⇒ q=9 (r/5)+1=(8/5) ⇒ r=3 Now f′(x)=−((3p(x^(1/3) +1)^(4/5) )/(5x^(8/5) )) ⇒ p=−(5/3) ⇒ ∫(((x^(1/3) +1)^(4/5) )/x^(8/5) )dx=−((5(x^(1/3) +1)^(9/5) )/(3x^(3/5) ))+C](https://www.tinkutara.com/question/Q200953.png)
$$\mathrm{Assume} \\ $$$$\int\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }{{x}^{\frac{\mathrm{8}}{\mathrm{5}}} }{dx}=\frac{{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{{q}}{\mathrm{5}}} }{{x}^{\frac{{r}}{\mathrm{5}}} } \\ $$$${f}'\left({x}\right)=\frac{{d}}{{dx}}\left[\frac{{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{{q}}{\mathrm{5}}} }{{x}^{\frac{{r}}{\mathrm{5}}} }\right]= \\ $$$$\frac{{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{{q}}{\mathrm{5}}−\mathrm{1}} \left(\left({q}−\mathrm{3}{r}\right){x}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{3}{r}\right)}{{x}^{\frac{{r}}{\mathrm{5}}+\mathrm{1}} } \\ $$$$\frac{{q}}{\mathrm{5}}−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\:{q}=\mathrm{9} \\ $$$$\frac{{r}}{\mathrm{5}}+\mathrm{1}=\frac{\mathrm{8}}{\mathrm{5}}\:\Rightarrow\:{r}=\mathrm{3} \\ $$$$\mathrm{Now}\:{f}'\left({x}\right)=−\frac{\mathrm{3}{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }{\mathrm{5}{x}^{\frac{\mathrm{8}}{\mathrm{5}}} }\:\Rightarrow\:{p}=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\int\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }{{x}^{\frac{\mathrm{8}}{\mathrm{5}}} }{dx}=−\frac{\mathrm{5}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{9}}{\mathrm{5}}} }{\mathrm{3}{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }+{C} \\ $$