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Indefinite Integrals

Section: Mathematics

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Question : 67 of 90

Marks: +1, -0

If ∫

\frac{\sqrt{1-x^{2}}}{x^{4}}

\left(\sqrt{1-x^{2}}\right)^{m}

+ C , for a suitable chosen integer m and a functionA(x), where C is a constant of integration then

(A(x))^{m}

[11 Jan 2019 Shift 1]

- $\frac{1}{3x^{3}}$
- $\frac{1}{27x^{9}}$
$\frac{1}{9x^{4}}$
$\frac{1}{27x^{6}}$

Solution:

∫

\frac{\lvert x\rvert \sqrt{\frac{1}{x^{2}}-1}}{x^{4}}

dx ;

Put

\frac{1}{x^{2}}

- 1 = t ⇒

\frac{dt}{dx}

= -

\frac{2}{x^{3}}

Case 1 : x ≥ 0

-

\frac{1}{2}\int\sqrt{t}

dt ⇒ -

\frac{t^{3/2}}{3}

+ C

⇒ -

\frac{1}{2}\left(\frac{1}{x^{2}}-1\right)^{3/2}

⇒

\frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{2}}

+ C

A (x) =

-\frac{1}{3x^{2}}

and m = 3

(A(x))^{m}

= -

\left(\frac{1}{3x^{3}}\right)^{3}

= -

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

Case-II x ≤ 0

We get

\frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{3}}

+ C

A (x) =

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

, m = 3

(A(m))^{m}

= -

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

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