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Mathematics Question on Integrals of Some Particular Functions

If ∫1−x2x4dx=A(x)(1−x2)m+C\int \frac{\sqrt{1-x^{2}}}{x^{4}} dx = A \left(x\right)\left(\sqrt{1-x^{2}}\right)^{m} + C , for a suitable chosen integer mm and a function A(x)A(x), where CC is a constant of integration then (A(x))m(A(x))^m equals :

A

−13x3\frac{-1}{3x^3}

B

−127x9\frac{-1}{27 x^9}

C

19x4\frac{1}{9 x^4}

D

127x6\frac{1}{27 x^6}

Answer

−127x9\frac{-1}{27 x^9}

Explanation

Solution

∫1−x2x4dx=A(x)(1−x2)m+C\int \frac{\sqrt{1-x^{2}}}{x^{4}} dx = A\left(x\right) \left(\sqrt{1-x^{2}}\right)^{m} +C
∫∣x∣1x2−1x4dx\int \frac{\left|x \right|\sqrt{\frac{1}{x^{2}} -1}}{x^{4}} dx
Put 1x2−1=t⇒dtdx=−2x3 \frac{1}{x^{2} } - 1 = t \Rightarrow \frac{dt}{dx} = \frac{-2}{x^{3}}
Case-1 x ≥\ge 0
−12∫tdt⇒−t3/23+C- \frac{1}{2} \int \sqrt{t} dt \Rightarrow - \frac{t^{3/2}}{3} + C
⇒−13(1x2−1)3/2\Rightarrow - \frac{1}{3} \left(\frac{1}{x^{2}}-1\right) ^{3/2}
⇒(1−x2)3−3x2+C\Rightarrow \frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{2}} +C
A(x)=−13x3A\left(x\right) = - \frac{1}{3x^{3}}
(A(x))m=(−13x3)3=−127x9\left(A\left(x\right)\right)^{m} = \left(- \frac{1}{3x^{3}}\right)^{3} = - \frac{1}{27x^{9}}
Case-II x ≤\le 0
We get (1−x2)3−3x3+C\frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{3}} +C
A(x)=1−3x3,m=3A\left(x\right) = \frac{1}{-3x^{3}} , m = 3
(A(x))m=−127x9\left(A\left(x\right)\right)^{m} = \frac{-1}{27x^{9}}