Let ∫ f(x) dx = F(x)+C . Consider the following statements: (A) ∫ f(3…

The correct answer is Option (2) → (A), (C) and (D) only

Since \(F'(x) = f(x)\), each statement is checked by differentiating the right-hand side with the chain rule:

    \( \displaystyle \frac{d}{dx}\left[\frac{1}{3}F(3x)\right] = \frac{1}{3}\times 3f(3x) = f(3x) \)

    \( \displaystyle \frac{d}{dx}\left[2F(2x+5)\right] = 2\times 2f(2x+5) = 4f(2x+5) \)

    \( \displaystyle \frac{d}{dx}\left[\frac{1}{2}F\left(x^2\right)\right] = \frac{1}{2}\times 2x\,f\left(x^2\right) = x\,f\left(x^2\right) \)

    \( \displaystyle \frac{d}{dx}\left[F(\sin x)\right] = \cos x\,f(\sin x) \)

So the right-hand sides of (A), (C) and (D) have exactly the given integrands as derivatives, while the derivative in (B) is \(4f(2x+5)\); the correct result is \(\int f(2x+5)\,dx = \frac{1}{2}F(2x+5)+C\).

Checking the statements

(A) \(\int f(3x)\,dx = \frac{1}{3}F(3x)+C\) — Correct. The derivative of \(\frac{1}{3}F(3x)\) is \(f(3x)\).

(B) \(\int f(2x+5)\,dx = 2F(2x+5)+C\) — Incorrect. The factor is \(\frac{1}{2}\), not 2; the derivative of \(2F(2x+5)\) is \(4f(2x+5)\).

(C) \(\int x\,f\left(x^2\right)dx = \frac{1}{2}F\left(x^2\right)+C\) — Correct. With \(t = x^2\), \(x\,dx = \frac{dt}{2}\).

(D) \(\int\cos x\,f(\sin x)\,dx = F(\sin x)+C\) — Correct. With \(t = \sin x\), \(dt = \cos x\,dx\).