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Step-by-Step Solution
Step 1: Rewrite the integral
We need to evaluate
$ \int \frac{\sqrt{1 - x^2}}{x^4} \, dx. $
The result is proposed to be of the form
$A(x)\bigl(\sqrt{1 - x^2}\bigr)^m + C,$
where $m$ is an integer and $A(x)$ is some function of $x$.
Step 2: Perform a suitable substitution
Observe that the integrand involves $ \sqrt{1 - x^2} $ and $x^4$ in the denominator. A natural approach is to consider expressions related to
$ \frac{1}{x^2} - 1, $
since it can transform the square root part neatly. Let us set:
$ t = \frac{1}{x^2} - 1. $
Now we compute $ \frac{dt}{dx} $:
$$
\frac{dt}{dx}
= \frac{d}{dx}\Bigl(\frac{1}{x^2} - 1\Bigr)
= \frac{d}{dx}\Bigl(\frac{1}{x^2}\Bigr)
- \frac{d}{dx}(1)
= -\frac{2}{x^3}.
$$
Thus,
$$
dt = -\frac{2}{x^3} \, dx
\quad\Longrightarrow\quad
dx = -\frac{x^3}{2} \, dt.
$$
Step 3: Express the integral in terms of t
First, note that
$$
1 - x^2
= x^2 \Bigl(\frac{1}{x^2} - 1\Bigr)
= x^2 t.
$$
Hence,
$$
\sqrt{1 - x^2}
= \sqrt{x^2 t}
= |x|\sqrt{t}.
$$
For the integral, we also have a factor $\frac{1}{x^4}$ from the original integrand:
$$
\frac{\sqrt{1 - x^2}}{x^4}
= \frac{|x|\sqrt{t}}{x^4}.
$$
When multiplying by $dx$, we use $dx = -\frac{x^3}{2}\,dt$, so:
$$
\int \frac{\sqrt{1 - x^2}}{x^4} \, dx
= \int \frac{|x|\sqrt{t}}{x^4}
\left( -\frac{x^3}{2}\,dt \right).
$$
This simplifies (assuming $x > 0$ for one case, then similarly for $x < 0$) to:
$$
= -\frac{1}{2} \int \frac{|x|\cdot x^3 \sqrt{t}}{x^4} \, dt
= -\frac{1}{2} \int \sqrt{t}\, dt.
$$
Step 4: Evaluate the integral in t
We now integrate
$ -\frac{1}{2} \int \sqrt{t}\, dt. $
Since
$ \sqrt{t} = t^{\frac{1}{2}}, $
we have
$$
\int t^{\frac{1}{2}} \, dt
= \frac{2}{3} t^{\frac{3}{2}}.
$$
Therefore,
$$
-\frac{1}{2} \int \sqrt{t}\, dt
= -\frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}}
= -\frac{1}{3} t^{\frac{3}{2}} + C.
$$
Step 5: Rewrite in terms of x
Recall
$ t = \frac{1}{x^2} - 1, $
so
$$
-\frac{1}{3} \left(\frac{1}{x^2} - 1\right)^{\frac{3}{2}} + C.
$$
Furthermore,
$$
\left(\frac{1}{x^2} - 1\right)^{\frac{3}{2}}
= \Bigl(\frac{1 - x^2}{x^2}\Bigr)^{\frac{3}{2}}
= \frac{(1 - x^2)^{\frac{3}{2}}}{|x|^3}.
$$
When $x > 0$, $|x| = x$, giving
$$
-\frac{1}{3} \left(\frac{1}{x^2} - 1\right)^{\frac{3}{2}}
= -\frac{(1 - x^2)^{\frac{3}{2}}}{3x^3}.
$$
Thus the integral becomes
$$
\int \frac{\sqrt{1 - x^2}}{x^4} \, dx
= -\frac{(1 - x^2)^{\frac{3}{2}}}{3x^3} + C.
$$
Step 6: Match the given form and identify A(x) and m
The result is expressed as
$ A(x)\bigl(\sqrt{1 - x^2}\bigr)^m + C. $
We see that
$ \bigl(\sqrt{1 - x^2}\bigr)^3 = (1 - x^2)^{\frac{3}{2}}, $
so comparing to
$ -\frac{(1 - x^2)^{\frac{3}{2}}}{3x^3}, $
we deduce:
$$
A(x) = -\frac{1}{3x^3}
\quad \text{and} \quad
m = 3.
$$
Step 7: Compute $(A(x))^m$
We want
$ (A(x))^m = \bigl(-\frac{1}{3x^3}\bigr)^3. $
Calculate:
$$
\bigl(-\frac{1}{3x^3}\bigr)^3
= -\frac{1}{27x^9}.
$$
Hence,
$$
(A(x))^m = -\frac{1}{27x^9}.
$$
Final Answer
The expression for $(A(x))^m$ is
$$
-\frac{1}{27 x^9}.
$$
