\(\int\displaystyle\frac{dx}{5+4x-x^2}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{6}\log\left|\frac{x+1}{5-x}\right|+C\)
Completing the square in the denominator:
\(5+4x-x^2 = 5-(x^2-4x+4)+4\)
\(= 9-(x-2)^2\)
With \(t = x-2\), this is the standard integral \(\int\frac{dt}{a^2-t^2} = \frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|+C\), with a = 3:
\( \displaystyle \int\frac{dx}{9-(x-2)^2} = \frac{1}{6}\log\left|\frac{3+(x-2)}{3-(x-2)}\right|+C \)
\( \displaystyle = \frac{1}{6}\log\left|\frac{x+1}{5-x}\right|+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{6}\log\left|\frac{5-x}{x+1}\right|+C\) — Incorrect. The fraction is upside down, so it is the negative of the answer.
Option (2) \(\displaystyle\frac{1}{3}\log\left|\frac{x+1}{5-x}\right|+C\) — Incorrect. The factor in front is \(\frac{1}{2a} = \frac{1}{6}\), not \(\frac{1}{a}\).
Option (3) \(\displaystyle\frac{1}{6}\log\left|\frac{x-1}{x+5}\right|+C\) — Incorrect. It comes from \((x+2)^2-9 = x^2+4x-5\), a different quadratic.