Question

(int left( \frac{3x^3 - 4x - 2}{x^3}
ight) dx = square + c)

Explanation:

Step1: Simplify the integrand

First, we simplify the fraction \(\frac{3x^{3}-4x - 2}{x^{3}}\) by dividing each term in the numerator by \(x^{3}\).
We know that \(\frac{a + b+ c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\), so \(\frac{3x^{3}-4x - 2}{x^{3}}=\frac{3x^{3}}{x^{3}}-\frac{4x}{x^{3}}-\frac{2}{x^{3}}\)
Simplifying each term: \(\frac{3x^{3}}{x^{3}} = 3\), \(\frac{4x}{x^{3}}=\frac{4}{x^{2}}=4x^{-2}\), \(\frac{2}{x^{3}} = 2x^{-3}\)
So the integrand becomes \(3-4x^{-2}-2x^{-3}\)

Step2: Integrate term - by - term

We use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int kdx=kx + C\) (where \(k\) is a constant)

• For the first term \(3\): \(\int3dx=3x\)
• For the second term \(-4x^{-2}\): \(\int-4x^{-2}dx=-4\times\frac{x^{-2 + 1}}{-2+1}=-4\times\frac{x^{-1}}{-1}=4x^{-1}=\frac{4}{x}\)
• For the third term \(-2x^{-3}\): \(\int-2x^{-3}dx=-2\times\frac{x^{-3+1}}{-3 + 1}=-2\times\frac{x^{-2}}{-2}=x^{-2}=\frac{1}{x^{2}}\)

Answer:

\(3x+\frac{4}{x}+\frac{1}{x^{2}}+C\)