∫(3x+2)/(x^2+6x+10) dx is equal to

The correct answer is Option (2) → \(\displaystyle\frac{3}{2}\log\left(x^2+6x+10\right)-7\tan^{-1}(x+3)+C\)

The derivative of \(x^2+6x+10\) is \(2x+6\). Writing the numerator as a multiple of this derivative plus a constant:

    \( \displaystyle 3x+2 = \frac{3}{2}(2x+6)-7 \)

So

    \( \displaystyle I = \frac{3}{2}\int\frac{2x+6}{x^2+6x+10}\,dx-7\int\frac{dx}{x^2+6x+10} \)

The first part is of the form \(\int\frac{f'(x)}{f(x)}\,dx = \log|f(x)|\):

    \( \displaystyle \frac{3}{2}\int\frac{2x+6}{x^2+6x+10}\,dx = \frac{3}{2}\log\left(x^2+6x+10\right) \)

For the second part, completing the square, \(x^2+6x+10 = (x+3)^2+1\):

    \( \displaystyle \int\frac{dx}{(x+3)^2+1} = \tan^{-1}(x+3) \)

Adding:

    \( \displaystyle I = \frac{3}{2}\log\left(x^2+6x+10\right)-7\tan^{-1}(x+3)+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{3}{2}\log\left(x^2+6x+10\right)+7\tan^{-1}(x+3)+C\) — Incorrect. The constant left over is \(2-9 = -7\), so the inverse tangent term is subtracted.

• Option (3) \(3\log\left(x^2+6x+10\right)-7\tan^{-1}(x+3)+C\) — Incorrect. The multiplier of \(2x+6\) is \(\frac{3}{2}\), not 3.

• Option (4) \(\displaystyle\frac{3}{2}\log\left(x^2+6x+10\right)-11\tan^{-1}(x+3)+C\) — Incorrect. The leftover constant is \(2-\frac{3}{2}\times 6 = -7\), not \(-11\).