Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 142318 by mnjuly1970 last updated on 29/May/21

Nice...≽≽≽∗∗∗≼≼≼...Calculus Ω:=∫_0 ^( 1) (((1−(x)^(1/3) )(1−((x ))^(1/5) )(1−(x)^(1/7) ))/(ln( ((x ))^(1/3) ))) dx=? ....m.n

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Nice}...\succcurlyeq\succcurlyeq\succcurlyeq\ast\ast\ast\preccurlyeq\preccurlyeq\preccurlyeq...{Calculus} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\:\right)\left(\mathrm{1}−\sqrt[{\mathrm{5}}]{{x}\:}\:\right)\left(\mathrm{1}−\sqrt[{\mathrm{7}}]{{x}}\:\right)}{{ln}\left(\:\sqrt[{\mathrm{3}}]{{x}\:\:}\:\right)}\:{dx}=? \\ $$$$\:\:\:\:\:\:\:....{m}.{n} \\ $$

Answered by Dwaipayan Shikari last updated on 29/May/21

K(a)=∫_0 ^1 (((1−x^a )(1−(x)^(1/5) )(1−(x)^(1/7) ))/(log(x)))dx K′(a)=−∫_0 ^1 x^a (1−(x)^(1/5) )(1−(x)^(1/7) )dx=−∫_0 ^1 x^a −x^(a+(1/5)) −x^(a+(1/7)) +x^(a+((12)/(35))) dx K′(a)=−(1/(a+1))+(1/(a+(6/5)))+(1/(a+(8/7)))−(1/(a+((47)/(35)))) K(a)=log(((a+(6/5))/(a+1)).((a+(8/7))/(a+((47)/(35)))))+C K(0)=0⇒C=log(((47)/(48))) K(a)=log(((a+(6/5))/(a+1)).((a+(8/7))/(a+((47)/(35)))).((47)/(48)))

$$\mathcal{K}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{{a}} \right)\left(\mathrm{1}−\sqrt[{\mathrm{5}}]{{x}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{7}}]{{x}}\right)}{{log}\left({x}\right)}{dx} \\ $$$$\mathcal{K}'\left({a}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \left(\mathrm{1}−\sqrt[{\mathrm{5}}]{{x}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{7}}]{{x}}\right){dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} −{x}^{{a}+\frac{\mathrm{1}}{\mathrm{5}}} −{x}^{{a}+\frac{\mathrm{1}}{\mathrm{7}}} +{x}^{{a}+\frac{\mathrm{12}}{\mathrm{35}}} {dx} \\ $$$$\mathcal{K}'\left({a}\right)=−\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{1}}{{a}+\frac{\mathrm{6}}{\mathrm{5}}}+\frac{\mathrm{1}}{{a}+\frac{\mathrm{8}}{\mathrm{7}}}−\frac{\mathrm{1}}{{a}+\frac{\mathrm{47}}{\mathrm{35}}} \\ $$$$\mathcal{K}\left({a}\right)={log}\left(\frac{{a}+\frac{\mathrm{6}}{\mathrm{5}}}{{a}+\mathrm{1}}.\frac{{a}+\frac{\mathrm{8}}{\mathrm{7}}}{{a}+\frac{\mathrm{47}}{\mathrm{35}}}\right)+{C} \\ $$$$\mathcal{K}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{C}={log}\left(\frac{\mathrm{47}}{\mathrm{48}}\right) \\ $$$$\mathcal{K}\left({a}\right)={log}\left(\frac{{a}+\frac{\mathrm{6}}{\mathrm{5}}}{{a}+\mathrm{1}}.\frac{{a}+\frac{\mathrm{8}}{\mathrm{7}}}{{a}+\frac{\mathrm{47}}{\mathrm{35}}}.\frac{\mathrm{47}}{\mathrm{48}}\right) \\ $$

Commented by mnjuly1970 last updated on 29/May/21

thank you so much...

$${thank}\:{you}\:{so}\:{much}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com