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 $\int \frac{\log (1+x^4)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$,then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

$\int \frac{x^2+1}{x^4+7 x^2+1} d x$ is equal to

If $f(x) = \int {\left( {\frac{{{x^2} + {{\sin }^2}x}}{{1 + {x^2}}}} \right)} {\sec ^2}x\,dx$ and $f(0) = 0,$ then $f(1)$ equals

$\int(\log (\sin x)+x \cot x) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module