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

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(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

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

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

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