Consider the following statements, where C is a constant of integration:
(A) \(\int\displaystyle\frac{2x-5}{x^2-5x+7}\,dx = \log\left(x^2-5x+7\right)+C\)
(B) \(\int\displaystyle\frac{dx}{x^2+x+1} = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C\)
(C) \(\int\displaystyle\frac{x}{x^2+4}\,dx = \frac{1}{2}\log\left(x^2+4\right)+C\)
(D) \(\int\displaystyle\frac{dx}{x^2-8x+15} = \frac{1}{2}\log\left|\frac{x-5}{x-3}\right|+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
In (A) and (C) the numerator is (a multiple of) the derivative of the denominator, so the f’/f form \(\int\frac{f'(x)}{f(x)}\,dx = \log|f(x)|\) applies:
\( \displaystyle \frac{d}{dx}\left(x^2-5x+7\right) = 2x-5,\qquad \frac{x}{x^2+4} = \frac{1}{2}\cdot\frac{2x}{x^2+4} \)
In (B) and (D), completing the square:
\( \displaystyle x^2+x+1 = \left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \displaystyle \int\frac{dx}{x^2+x+1} = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \)
\( x^2-8x+15 = (x-4)^2-1^2 \)
\( \displaystyle \int\frac{dx}{(x-4)^2-1} = \frac{1}{2}\log\left|\frac{x-5}{x-3}\right| \)
So (A), (C) and (D) are right, while the factor in (B) should be \(\frac{2}{\sqrt{3}}\).
Checking the statements
(A) \(\int\frac{2x-5}{x^2-5x+7}\,dx = \log\left(x^2-5x+7\right)+C\) — Correct. The numerator is exactly the derivative of the denominator.
(B) \(\int\frac{dx}{x^2+x+1} = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C\) — Incorrect. Here \(a = \frac{\sqrt{3}}{2}\), so the factor \(\frac{1}{a}\) is \(\frac{2}{\sqrt{3}}\).
(C) \(\int\frac{x}{x^2+4}\,dx = \frac{1}{2}\log\left(x^2+4\right)+C\) — Correct. The numerator x is half the derivative 2x.
(D) \(\int\frac{dx}{x^2-8x+15} = \frac{1}{2}\log\left|\frac{x-5}{x-3}\right|+C\) — Correct. With \(t = x-4\) and a = 1, \(\frac{1}{2a}\log\left|\frac{t-a}{t+a}\right|\) gives this.