int {frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}};d | vedclass

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

Question

(A) We have $I = \int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx}$.Taking $x^4$ common from the bracket,we get $I = \int {\frac{{{{[x^4(1 - 1/x^3)]}

Vedclass · Accepted Answer

$\frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$

Vedclass · Answer

(A) We have $I = \int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\,dx}$.Taking $x^4$ common from the bracket,we get $I = \int {\frac{{{{[x^4(1 - 1/x^3)]}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{x{{(1 - 1/x^3)}^{1/4}}}}{{{x^5}}}\,dx} = \int {\frac{{{{(1 - 1/x^3)}^{1/4}}}}{{{x^4}}}\,dx}$.Let $1 - \frac{1}{{{x^3}}} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{3}{{{x^4}}}dx = dt$,which implies $\frac{1}{{{x^4}}}dx = \frac{1}{3}dt$.Substituting these into the integral,we get $I = \int {{t^{1/4}} \cdot \frac{1}{3}dt} = \frac{1}{3} \cdot \frac{{{t^{5/4}}}}{{5/4}} + c = \frac{1}{3} \cdot \frac{4}{5} {t^{5/4}} + c = \frac{4}{{15}}{t^{5/4}} + c$.Substituting back $t = 1 - \frac{1}{{{x^3}}}$,we get $I = \frac{4}{{15}}{\left( {1 - \frac{1}{{{x^3}}}} \right)^{5/4}} + c$.

$\int {\frac{{{{({x^4} - x)}^{1/4}}}}{{{x^5}}}\;dx} $ is equal to

Similar Questions

$\int \frac{2+\cos \frac{x}{2}}{x+\sin \frac{x}{2}} \,d x=$

$\int \frac{d x}{x\left(x^4+1\right)}=$

$f(x) = \int \frac{dx}{\sin^6 x}$ is a polynomial of degree

The integral $\int \frac{\sec^2 x}{(\sec x+\tan x)^{9/2}} dx$ equals (for some arbitrary constant $K$):

If $c$ is any arbitrary constant,then $\int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module