\( \int\displaystyle\frac{1}{x^2+4x+8}\,dx \) is equal to (Where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{2}\tan^{-1}\displaystyle\frac{x+2}{2}+\mathrm{C} \)
Completing the square, \( x^2+4x+8=(x+2)^2+2^2 \). Using \( \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\frac{u}{a} \):
\( \displaystyle \int\frac{dx}{(x+2)^2+2^2}=\frac{1}{2}\tan^{-1}\frac{x+2}{2}+C \)
Why the other options are incorrect
Option (1) \( -2\ \tan^{-1}(x+2) \) — Incorrect. The factor should be \( \frac{1}{2} \) and the argument \( \frac{x+2}{2} \).
Option (2) \( \tan^{-1}\displaystyle\frac{x+2}{2}+\mathrm{C} \) — Incorrect. This misses the factor \( \frac{1}{a}=\frac{1}{2} \).
Option (3) \( -\displaystyle\frac{1}{2}\tan^{-1}\displaystyle\frac{x+2}{2}+\mathrm{C} \) — Incorrect. The sign is wrong: the integrand is positive, so the answer must increase with x.