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Give answer! - If for a suitable chosen integer m and a function , where C is a constant of integration , then - Integral Calculus - JEE Main

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}}

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A admin

 

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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