Consider the following statements, where C is a constant of integration and \(x \gt 0\):
(A) \(\int\left(2x\cos x-x^2\sin x\right)dx = x^2\cos x+C\)
(B) \(\int\displaystyle\frac{\sin x-x\cos x}{x^2}\,dx = \frac{\sin x}{x}+C\)
(C) \(\int\left(1+\log x\right)dx = x\log x+C\)
(D) \(\int\left(\tan x+x\sec^2x\right)dx = x\tan x+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Integration is the inverse process of differentiation, so each result is checked by differentiating its right-hand side with the product or quotient rule:
\( \displaystyle \frac{d}{dx}\left(x^2\cos x\right) = 2x\cos x-x^2\sin x \)
\( \displaystyle \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x\cos x-\sin x}{x^2} \)
\( \displaystyle \frac{d}{dx}\left(x\log x\right) = \log x+1 \)
\( \displaystyle \frac{d}{dx}\left(x\tan x\right) = \tan x+x\sec^2x \)
So (A), (C) and (D) are true, while \(\int\frac{\sin x-x\cos x}{x^2}\,dx = -\frac{\sin x}{x}+C\).
Checking the statements
(A) \(\int\left(2x\cos x-x^2\sin x\right)dx = x^2\cos x+C\) — Correct. The derivative of \(x^2\cos x\) is \(2x\cos x-x^2\sin x\).
(B) \(\int\frac{\sin x-x\cos x}{x^2}\,dx = \frac{\sin x}{x}+C\) — Incorrect. The derivative of \(\frac{\sin x}{x}\) is \(\frac{x\cos x-\sin x}{x^2}\), the negative of the integrand.
(C) \(\int\left(1+\log x\right)dx = x\log x+C\) — Correct. The derivative of \(x\log x\) is \(\log x+1\).
(D) \(\int\left(\tan x+x\sec^2x\right)dx = x\tan x+C\) — Correct. The derivative of \(x\tan x\) is \(\tan x+x\sec^2x\).