Question Number 141691 by mnjuly1970 last updated on 22/May/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{Calculus}\left({I}\right)…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\frac{\mathrm{1}}{\mathrm{2}\:}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} }{dx}=??? \\ $$
Answered by Dwaipayan Shikari last updated on 22/May/21
![∫_(1/2) ^1 (1/(x^5 (1+(1/x^4 ))^(3/4) ))dx =−(1/4)∫_(1/2) ^1 ((−(4/x^5 ) )/((1+(1/x^4 ))^(3/4) ))dx=[(1+(1/x^4 ))^(1/4) ]_1 ^(1/2) =(((17))^(1/4) −(2)^(1/4) )](https://www.tinkutara.com/question/Q141696.png)
$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{3}/\mathrm{4}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{−\frac{\mathrm{4}}{{x}^{\mathrm{5}} }\:}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{3}/\mathrm{4}} }{dx}=\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{1}/\mathrm{4}} \right]_{\mathrm{1}} ^{\mathrm{1}/\mathrm{2}} =\left(\sqrt[{\mathrm{4}}]{\mathrm{17}}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right) \\ $$