∫ x 2^x dx is equal to

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

Using integration by parts with x as the first function and \(2^x\) as the second (\(\int 2^x\,dx = \frac{2^x}{\log 2}\)):

    \( \displaystyle \int x\,2^x\,dx = \frac{x\,2^x}{\log 2}-\int\frac{2^x}{\log 2}\,dx \)

    \( \displaystyle = \frac{x\,2^x}{\log 2}-\frac{2^x}{(\log 2)^2}+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{x\,2^x}{\log 2}+\frac{2^x}{(\log 2)^2}+C\) — Incorrect. The by-parts formula subtracts the second integral, so the sign of the last term is wrong.

• Option (2) \(x\,2^x\log 2-2^x(\log 2)^2+C\) — Incorrect. It multiplies by \(\log 2\) where it should divide; \(\int 2^x\,dx = \frac{2^x}{\log 2}\).

• Option (3) \(\displaystyle\frac{2^x(x-1)}{\log 2}+C\) — Incorrect. It copies the result for \(xe^x\); the second term must be divided by \((\log 2)^2\).