∫(x^3+5x^2-4)/(x^2) dx is equal to

The correct answer is Option (4) → \(\displaystyle\frac{x^2}{2}+5x+\frac{4}{x}+C\)

Dividing each term of the numerator by \(x^2\):

    \( \displaystyle \frac{x^3+5x^2-4}{x^2} = x+5-4x^{-2} \)

Integrating each power with \(\int x^n\,dx = \frac{x^{n+1}}{n+1}\):

    \( \displaystyle \int\left(x+5-4x^{-2}\right)dx = \frac{x^2}{2}+5x-4\times\frac{x^{-1}}{-1}+C \)

    \( \displaystyle = \frac{x^2}{2}+5x+\frac{4}{x}+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{x^2}{2}+5x-\frac{4}{x}+C\) — Incorrect. The sign slip: \(\int x^{-2}\,dx = -x^{-1}\), so \(-4\) times it is \(+\frac{4}{x}\).

• Option (2) \(\displaystyle x^2+5x+\frac{4}{x}+C\) — Incorrect. The integral of x is \(\frac{x^2}{2}\), not \(x^2\).

• Option (3) \(\displaystyle\frac{x^2}{2}+5x-4\log|x|+C\) — Incorrect. The log rule applies only to \(x^{-1}\); here the power is \(x^{-2}\).