Integrate \(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\).
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Correct Answer: Option A
Explanation:
\(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\)
= \(\frac{4x^{-3 + 1}}{-3 + 1} - \frac{7x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} - 6x\)
= \(-2x^{-2} - \frac{7}{3} x^3 + \frac{5}{2} x^2 - 6x\)
\(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\)
= \(\frac{4x^{-3 + 1}}{-3 + 1} - \frac{7x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} - 6x\)
= \(-2x^{-2} - \frac{7}{3} x^3 + \frac{5}{2} x^2 - 6x\)