int_{frac{1}{sqrt[5]{31}}}^{frac{1}{sqrt[5]{2 | vedclass

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(D) माना $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}+x^{25}}}$.हर से $x^{30}$ कॉमन लेने पर: $I = \int_{\fra

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$\frac{-65}{4}$

Vedclass · Answer

(D) माना $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}+x^{25}}}$.हर से $x^{30}$ कॉमन लेने पर: $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}(1+x^{-5})}} = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{x^6(1+x^{-5})^{1/5}}$.माना $t = 1 + x^{-5}$. तब $dt = -5x^{-6} dx$,जिसका अर्थ है $x^{-6} dx = -\frac{1}{5} dt$.समाकलन की सीमाएं बदलने पर:जब $x = \frac{1}{\sqrt[5]{31}}$,तब $t = 1 + (\sqrt[5]{31})^5 = 1 + 31 = 32$.जब $x = \frac{1}{\sqrt[5]{242}}$,तब $t = 1 + (\sqrt[5]{242})^5 = 1 + 242 = 243$.इन मानों को समाकलन में रखने पर:$I = \int_{32}^{243} -\frac{1}{5} t^{-1/5} dt = -\frac{1}{5} \left[ \frac{t^{4/5}}{4/5} \right]_{32}^{243} = -\frac{1}{4} [t^{4/5}]_{32}^{243}$.$I = -\frac{1}{4} (243^{4/5} - 32^{4/5}) = -\frac{1}{4} ((3^5)^{4/5} - (2^5)^{4/5}) = -\frac{1}{4} (3^4 - 2^4) = -\frac{1}{4} (81 - 16) = -\frac{65}{4}$.

$\int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30}+x^{25}}} d x=$

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