If int frac{sqrt{1-x^2}}{x^4} ,dx = A(x)left( | vedclass

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

Question

(C) Let $I = \int \frac{\sqrt{1-x^2}}{x^4} \,dx$.We can rewrite the integral as $I = \int \frac{x \sqrt{\frac{1}{x^2}-1}}{x^4} \,dx = \int \frac{\sqrt

Vedclass · Accepted Answer

$\frac{-1}{27 x^9}$

Vedclass · Answer

(C) Let $I = \int \frac{\sqrt{1-x^2}}{x^4} \,dx$.We can rewrite the integral as $I = \int \frac{x \sqrt{\frac{1}{x^2}-1}}{x^4} \,dx = \int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3} \,dx$.Let $\frac{1}{x^2}-1 = t$. Then $-\frac{2}{x^3} \,dx = dt$,which implies $\frac{1}{x^3} \,dx = -\frac{dt}{2}$.Substituting these into the integral,we get $I = -\frac{1}{2} \int \sqrt{t} \,dt = -\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + c = -\frac{1}{3} t^{3/2} + c$.Substituting $t = \frac{1-x^2}{x^2}$ back,we get $I = -\frac{1}{3} \left(\frac{1-x^2}{x^2}\right)^{3/2} + c = -\frac{1}{3} \frac{(\sqrt{1-x^2})^3}{(x^2)^{3/2}} + c = -\frac{1}{3x^3} (\sqrt{1-x^2})^3 + c$.Comparing this with $A(x)(\sqrt{1-x^2})^m + c$,we identify $A(x) = -\frac{1}{3x^3}$ and $m = 3$.Therefore,$(A(x))^m = \left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}$.

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

Similar Questions

$\int \frac{\sec^2 x}{(\sec x + \tan x)^{5/2}} dx =$

Let $f(x) = \int \frac{x^2 dx}{(1 + x^2)(1 + \sqrt{1 + x^2})}$ and $f(0) = 0$,then the value of $f(1)$ is:

The value of $\int \frac{x^3}{\sqrt{1 + x^4}} \, dx$ is

The integral $\int \frac{2x^3 - 1}{x^4 + x} \,dx$ is equal to (Here $C$ is a constant of integration)

$\int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module