\(\displaystyle\int \frac{(x + 1)^2}{x}\,dx,\ x \gt 0\), is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x^2}{2} + 2x + \log x + C\)
Expanding the square and dividing each term by x:
\( \displaystyle \frac{(x + 1)^2}{x} = \frac{x^2 + 2x + 1}{x} = x + 2 + \frac{1}{x} \)
Integrating term by term:
\( \displaystyle \int \left(x + 2 + \frac{1}{x}\right)dx = \frac{x^2}{2} + 2x + \log x + C \)
(Since x > 0, \(\log|x| = \log x\).)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{(x + 1)^3}{3x} + C\) — Incorrect. This integrates the numerator and keeps the denominator x unchanged; x is not a constant, so it cannot be kept outside.
Option (2) \(\displaystyle\frac{x^2}{2} + 2x + C\) — Incorrect. The term \(\frac{1}{x}\) has been left out; its integral is \(\log x\).
Option (3) \(\displaystyle\frac{x^2}{2} + x + \log x + C\) — Incorrect. The middle term of \((x + 1)^2\) is 2x, so after division the constant is 2 and its integral is 2x, not x.