Consider the following statements, where C is a constant of integration:
(A) \(\int \log x\,dx = x\log x - x + C\)
(B) \(\int\displaystyle\frac{dx}{x^2 + 16} = \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) + C\)
(C) \(\int 3^x\,dx = 3^x\log 3 + C\)
(D) \(\int |x|\,dx = \displaystyle\frac{1}{2}x|x| + C\), \(x \ne 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
A result \(\int f(x)\,dx = F(x) + C\) is correct exactly when \(F'(x) = f(x)\). Differentiating each right-hand side:
\( \displaystyle \frac{d}{dx}(x\log x - x) = \log x + 1 - 1 = \log x \)
\( \displaystyle \frac{d}{dx}\left(\frac{1}{4}\tan^{-1}\frac{x}{4}\right) = \frac{1}{4} \cdot \frac{1/4}{1 + x^2/16} = \frac{1}{x^2 + 16} \)
\( \displaystyle \frac{d}{dx}\left(3^x\log 3\right) = 3^x(\log 3)^2 \)
For \(x \gt 0\), \(\frac{1}{2}x|x| = \frac{x^2}{2}\) has derivative \(x = |x|\); for \(x \lt 0\), \(\frac{1}{2}x|x| = -\frac{x^2}{2}\) has derivative \(-x = |x|\).
So the derivatives are \(\log x\), \(\frac{1}{x^2 + 16}\), \(3^x(\log 3)^2\) and \(|x|\), while the correct result for (C) is \(\int 3^x\,dx = \frac{3^x}{\log 3} + C\).
Checking the statements
(A) \(\int \log x\,dx = x\log x - x + C\) — Correct. The derivative of \(x\log x - x\) is \(\log x\).
(B) \(\int \frac{dx}{x^2 + 16} = \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) + C\) — Correct. This is \(\frac{1}{a}\tan^{-1}\frac{x}{a}\) with \(a = 4\).
(C) \(\int 3^x\,dx = 3^x\log 3 + C\) — Incorrect. One must divide by \(\log 3\): the answer is \(\frac{3^x}{\log 3} + C\).
(D) \(\int |x|\,dx = \frac{1}{2}x|x| + C\), \(x \ne 0\) — Correct. The derivative of \(\frac{1}{2}x|x|\) is \(|x|\) on both sides of 0.