int frac{1}{x^5 sqrt[5]{x^5+1}} d x= | vedclass

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(C) Let $I = \int \frac{1}{x^5 \sqrt[5]{x^5+1}} dx$. Factor out $x^5$ from the radical: $\sqrt[5]{x^5+1} = \sqrt[5]{x^5(1 + x^{-5})} = x(1 + x^{-5})^{

Vedclass · Accepted Answer

$-\frac{\left(x^5+1\right)^{\frac{4}{5}}}{4 x^4}+c$

Vedclass · Answer

(C) Let $I = \int \frac{1}{x^5 \sqrt[5]{x^5+1}} dx$. Factor out $x^5$ from the radical: $\sqrt[5]{x^5+1} = \sqrt[5]{x^5(1 + x^{-5})} = x(1 + x^{-5})^{1/5}$. Substituting this into the integral: $I = \int \frac{1}{x^5 \cdot x(1 + x^{-5})^{1/5}} dx = \int \frac{1}{x^6 (1 + x^{-5})^{1/5}} dx$. Let $t = 1 + x^{-5}$. Then $dt = -5x^{-6} dx$,which implies $x^{-6} dx = -\frac{1}{5} dt$. Substituting $t$ into the integral: $I = \int \frac{-1/5}{t^{1/5}} dt = -\frac{1}{5} \int t^{-1/5} dt$. Integrating: $I = -\frac{1}{5} \cdot \frac{t^{4/5}}{4/5} + C = -\frac{1}{4} t^{4/5} + C$. Substituting back $t = 1 + x^{-5} = \frac{x^5+1}{x^5}$: $I = -\frac{1}{4} \left(\frac{x^5+1}{x^5}\right)^{4/5} + C = -\frac{(x^5+1)^{4/5}}{4(x^5)^{4/5}} + C = -\frac{(x^5+1)^{4/5}}{4x^4} + C$.

$\int \frac{1}{x^5 \sqrt[5]{x^5+1}} d x=$

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