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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}+x^{25}}}$.Factor out $x^{30}$ from the denominator: $I

Vedclass · Accepted Answer

$\frac{-65}{4}$

Vedclass · Answer

(D) Let $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}+x^{25}}}$.Factor out $x^{30}$ from the denominator: $I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{\sqrt[5]{x^{30}(1+x^{-5})}} = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{dx}{x^6(1+x^{-5})^{1/5}}$.Let $t = 1 + x^{-5}$. Then $dt = -5x^{-6} dx$,which implies $x^{-6} dx = -\frac{1}{5} dt$.Change the limits of integration:When $x = \frac{1}{\sqrt[5]{31}}$,$t = 1 + (\sqrt[5]{31})^5 = 1 + 31 = 32$.When $x = \frac{1}{\sqrt[5]{242}}$,$t = 1 + (\sqrt[5]{242})^5 = 1 + 242 = 243$.Substituting these into the integral:$I = \int_{32}^{243} -\frac{1}{5} t^{-1/5} dt = -\frac{1}{5} \left[ \frac{t^{4/5}}{4/5} \right]_{32}^{243} = -\frac{1}{4} [t^{4/5}]_{32}^{243}$.$I = -\frac{1}{4} (243^{4/5} - 32^{4/5}) = -\frac{1}{4} ((3^5)^{4/5} - (2^5)^{4/5}) = -\frac{1}{4} (3^4 - 2^4) = -\frac{1}{4} (81 - 16) = -\frac{65}{4}$.

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

Similar Questions

If $\int \frac{1 - (\cot x)^{2021}}{\tan x + (\cot x)^{2022}} dx = \frac{1}{A} \log |(\sin x)^{2023} + (\cos x)^{2023}| + c$,then $A = . . . . . .$

Let $I_n = \int \tan^n x dx, (n > 1)$. If $I_4 + I_6 = a \tan^5 x + b x^5 + C$,where $C$ is the constant of integration,then the ordered pair $(a, b)$ is equal to:

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$

$\int \frac{1}{16-7 \sin ^2 x} d x=$

$\int \frac{d x}{\sin x+\cos x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module