∫(x+2)/(2x^2+6x+5) dx is equal to (C is a constant of integration)

The correct answer is Option (2) → \(\displaystyle\frac{1}{4}\log\left(2x^2+6x+5\right)+\frac{1}{2}\tan^{-1}(2x+3)+C\)

The derivative of the denominator is 4x + 6, so the numerator is written as a multiple of the derivative plus a constant:

    \( \displaystyle x+2 = \frac{1}{4}(4x+6)+\frac{1}{2} \)

So

    \( \displaystyle I = \frac{1}{4}\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac{1}{2}\int\frac{dx}{2x^2+6x+5} \)

The first part is of the form \(\int\frac{f'(x)}{f(x)}\,dx\), and \(2x^2+6x+5 \gt 0\) for all x:

    \( \displaystyle \frac{1}{4}\int\frac{4x+6}{2x^2+6x+5}\,dx = \frac{1}{4}\log\left(2x^2+6x+5\right) \)

For the second part, take out 2 and complete the square:

    \( \displaystyle 2x^2+6x+5 = 2\left[\left(x+\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2\right] \)

    \( \displaystyle \int\frac{dx}{2x^2+6x+5} = \frac{1}{2}\times 2\tan^{-1}\left(\frac{x+\frac{3}{2}}{\frac{1}{2}}\right) \)

    \( = \tan^{-1}(2x+3) \)

Hence

    \( \displaystyle I = \frac{1}{4}\log\left(2x^2+6x+5\right)+\frac{1}{2}\tan^{-1}(2x+3)+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{1}{2}\log\left(2x^2+6x+5\right)+\frac{1}{2}\tan^{-1}(2x+3)+C\) — Incorrect. The derivative of the denominator is 4x + 6, so the log coefficient is \(\frac{1}{4}\), not \(\frac{1}{2}\).

• Option (3) \(\displaystyle\frac{1}{2}\log\left(2x^2+6x+5\right)+\frac{1}{4}\tan^{-1}(2x+3)+C\) — Incorrect. Both coefficients are wrong; the log part is \(\frac{1}{4}\) and the \(\tan^{-1}\) part is \(\frac{1}{2}\).

• Option (4) \(\log\left(2x^2+6x+5\right)+\displaystyle\frac{1}{2}\tan^{-1}(2x+3)+C\) — Incorrect. Its derivative has numerator \(4x+\frac{13}{2}\) over \(2x^2+6x+5\), not \(x+2\).