Question

\(\int\frac{1}{x^3}\sqrt{1-\frac{1}{x^2}}dx=\)

• A. \(\frac{-1}{6}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)
• B. \(\frac{1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)
• C. \(\frac{-1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)
• D. \(\frac{4}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)
• E. \(\frac{-4}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)

Answer 1

The Correct Option is B

We are given the integral \[ I = \int \frac{1}{x^3} \sqrt{1 - \frac{1}{x^2}} \, dx. \] First, simplify the square root expression: \[ \sqrt{1 - \frac{1}{x^2}} = \left( 1 - \frac{1}{x^2} \right)^{\frac{1}{2}}. \] Thus, the integral becomes: \[ I = \int \frac{1}{x^3} \left( 1 - \frac{1}{x^2} \right)^{\frac{1}{2}} \, dx. \] Now, let us make the substitution \( u = 1 - \frac{1}{x^2} \), so that: \[ du = \frac{2}{x^3} \, dx. \] Rewriting the integral in terms of \( u \), we get: \[ I = \frac{1}{3} \int u^{\frac{1}{2}} \, du. \] Now, integrate \( u^{\frac{1}{2}} \): \[ I = \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{1}{3} \left( 1 - \frac{1}{x^2} \right)^{\frac{3}{2}} + C. \]

The correct option is (B) : \(\frac{1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C\)

Answer 2

We are asked to evaluate the integral: \[\int \frac{1}{x^3} \sqrt{1 - \frac{1}{x^2}} \, dx\]

We will use substitution. Let \(u = 1 - \frac{1}{x^2} = 1 - x^{-2}\). Then, the derivative of \(u\) with respect to \(x\) is: \[\frac{du}{dx} = 2x^{-3} = \frac{2}{x^3}\]

So, \(dx = \frac{x^3}{2} \, du\). We substitute \(u\) and \(dx\) into the integral: \[\int \frac{1}{x^3} \sqrt{1 - \frac{1}{x^2}} \, dx = \int \frac{1}{x^3} \sqrt{u} \cdot \frac{x^3}{2} \, du = \int \frac{1}{2}\sqrt{u} \, du\]

We rewrite \(\sqrt{u}\) as \(u^{1/2}\) and integrate with respect to \(u\): \[\int \frac{1}{2} u^{1/2} \, du = \frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{\frac{3}{2}} + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C\]

Now, substitute back \(u = 1 - \frac{1}{x^2}\): \[\frac{1}{3} \left(1 - \frac{1}{x^2}\right)^{3/2} + C\]

Therefore, the integral is: \[\int \frac{1}{x^3} \sqrt{1 - \frac{1}{x^2}} \, dx = \frac{1}{3} \left(1 - \frac{1}{x^2}\right)^{3/2} + C\]