\(\int\left(1+2x+3x^{2}+4x^{3}+\cdots\right)dx\), where \(|x| \lt 1\), is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left(1-x\right)^{-1}+C\)
Integrating term by term:
\( \int\left(1+2x+3x^{2}+4x^{3}+\cdots\right)dx = x+x^2+x^3+x^4+\cdots+C \)
For \(|x| \lt 1\) this is an infinite geometric series with first term x and ratio x:
\( \displaystyle x+x^2+x^3+\cdots = \frac{x}{1-x} \)
\( \displaystyle = \frac{1}{1-x}-1 \)
The −1 joins the constant, so the integral is
\( \left(1-x\right)^{-1}+C \)
Why the other options are incorrect
Option (1) \(\left(1+x\right)^{-1}+C\) — Incorrect. Its derivative is \(-\left(1+x\right)^{-2} = -1+2x-3x^2+\cdots\), with alternating signs.
Option (2) \(\left(1-x\right)^{-2}+C\) — Incorrect. \(\left(1-x\right)^{-2}\) is the sum of the integrand itself, not of its integral.
Option (4) \(\log\left(1-x\right)+C\) — Incorrect. Its derivative is \(-\frac{1}{1-x} = -\left(1+x+x^2+\cdots\right)\).