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

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

Question

(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 $t = 1 + x^{-5} = \frac{x^5+1}{x^5}$: $I = -\frac{1}{4} \left(\frac{x^5+1}{x^5}\right)^{4/5} + C = -\frac{(x^5+1)^{4/5}}{4(x^5)^{4/5}} + C = -\frac{(x^5+1)^{4/5}}{4x^4} + C$.

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

Similar Questions

If $\int e^{\sin ^2 x}(\sin x \cos x+\cos ^3 x \sin x) d x = e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx =$

If $\int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}} = \frac{-(1+x^m)^{n/m}}{nx^n} + C$,then $m-n=$

$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx = $ (where $C$ is a constant of integration)

$\int \sec x \tan^3 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module