\(\int\displaystyle\frac{x^2}{x+1}\,dx\) is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{x^2}{2}-x+\log|x+1|+C\)
The degree of the numerator is more than that of the denominator, so we divide first:
\( x^2 = (x+1)(x-1)+1 \)
\( \displaystyle \frac{x^2}{x+1} = x-1+\frac{1}{x+1} \)
Integrating term by term:
\( \displaystyle \int\frac{x^2}{x+1}\,dx = \frac{x^2}{2}-x+\log|x+1|+C \)