\( \int\displaystyle\frac{x^3}{x+2}dx \) is equal to: (where \( C \) is an arbitrary constant and \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{x^3}{3}-x^2+4x-8\log|x+2|+c \)
Since the numerator has the higher degree, \( x^3 \) is first divided by \( x+2 \). As \( x^3+8=(x+2)(x^2-2x+4) \),
\( \displaystyle \frac{x^3}{x+2}=\frac{(x^3+8)-8}{x+2} \)
\( \displaystyle =x^2-2x+4-\frac{8}{x+2} \)
Integrating term by term:
\( \displaystyle \int\frac{x^3}{x+2}dx=\frac{x^3}{3}-x^2+4x-8\log|x+2|+c \)
Shortcut: Differentiating Option (1) gives
\( \displaystyle x^2-2x+4-\frac{8}{x+2}=\frac{(x^2-2x+4)(x+2)-8}{x+2} \)
\( \displaystyle =\frac{x^3+8-8}{x+2}=\frac{x^3}{x+2} \)