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

then $(A(x))^{m}$ equals:
Option 1)

$\frac{1}{27x^{6}}$
Option 2)

$\frac{1}{9x^{4}}$
Option 3)

$\frac{-1}{27x^{9}}$
Option 4)

$\frac{-1}{3x^{3}}$

Answers (1)

best_answer

Extended forms of fundamental formulae -

If $x$ is replaced by a LINEAR FUNCTION of $x\Rightarrow \left ( ax+b \right )$ form then ,

$\int f\left ( ax+b \right )dx =\frac{F\left ( ax+b \right )}{\frac{\mathrm{d} }{\mathrm{d} x}\left ( ax+b \right )}+c$

- wherein

Fundamental formulae such as $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ , $\int sinx dx=-cosx$ ,..... and so on

Type of integration by substitution -

$\int (f(x))^{n}\cdot f{}'(x)dx$

$\therefore \frac{\left [ f(x) \right ]^{n+1}}{n+1}+c$

- wherein

Let $f(x)=t$

$f{}'(x)dx=dt$

$I = \int \frac{\sqrt{1-x^{2}}}{x^{4}}dx$ = $= \int \frac{1}{x^{3}}\sqrt{\frac{1}{x^{2}}-1}dx,$

substitute $\frac{1}{x^{2}}-1=t^{2}\Rightarrow \frac{dx}{x^{3}}=-tdt$

Hence, $I=-\int t^{2}dt$

$I=-\frac{t^{3}}{3}+c$

$=-\frac{1}{3}\left ( \frac{1}{x^{2}}-1 \right )^{3/2}+C$

$=-\frac{(\sqrt{1-x^{2}})}{3x^{3}}+c$

$\Rightarrow A(x)=-\frac{1}{3x^{3}}$ m=3

$\Rightarrow (A(x))^{3}=-\frac{1}{27x^{9}}$

Option 1)

$\frac{1}{27x^{6}}$

Option 2)

$\frac{1}{9x^{4}}$

Option 3)

$\frac{-1}{27x^{9}}$
Option 4)

$\frac{-1}{3x^{3}}$

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