Evaluate ∫ (4/x^3) dx.

Evaluate ∫ (4/x^3) dx.

• A. -2/x^2 + C
• B. 2/x^2 + C
• C. 4/x^2 + C
• D. -x^2 + C

Answer: (A)

To evaluate the integral of \(\frac{4}{x^3}\) with respect to \(x\), we can rewrite the integrand in a more manageable form. The expression \(\frac{4}{x^3}\) can be expressed as \(4x^{-3}\). Now, we can apply the power rule of integration, which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\). Here, \(n = -3\), so we apply the power rule: \[ \int 4x^{-3} \, dx = 4 \cdot \frac{x^{-3 + 1}}{-3 + 1} + C = 4 \cdot \frac{x^{-2}}{-2} + C \] This simplifies to: \[ 4 \cdot \left(-\frac{1}{2} x^{-2}\right) + C = -2x^{-2} + C \] Reverting \(x^{-2}\) back to its original form \( \frac{1}{x^2} \), we find:

To evaluate the integral of (\frac{4}{x^3}) with respect to (x), we can rewrite the integrand in a more manageable form. The expression (\frac{4}{x^3}) can be expressed as (4x^{-3}).

Now, we can apply the power rule of integration, which states that (\int x^n , dx = \frac{x^{n+1}}{n+1} + C) for (n \neq -1).

Here, (n = -3), so we apply the power rule:

[

\int 4x^{-3} , dx = 4 \cdot \frac{x^{-3 + 1}}{-3 + 1} + C = 4 \cdot \frac{x^{-2}}{-2} + C

]

This simplifies to:

[

4 \cdot \left(-\frac{1}{2} x^{-2}\right) + C = -2x^{-2} + C

]

Reverting (x^{-2}) back to its original form ( \frac{1}{x^2} ), we find: