\(\int\displaystyle\frac{x+3}{\sqrt{5-4x-x^2}}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle -\sqrt{5-4x-x^2}+\sin^{-1}\left(\frac{x+2}{3}\right)+C\)
The derivative of \(5-4x-x^2\) is \(-4-2x\). Writing the numerator as a multiple of this derivative plus a constant:
\( \displaystyle x+3 = -\frac{1}{2}(-4-2x)+1 \)
So the integral splits into two parts:
\( \displaystyle I = -\frac{1}{2}\int\frac{-4-2x}{\sqrt{5-4x-x^2}}\,dx+\int\frac{dx}{\sqrt{5-4x-x^2}} \)
For the first part, with \(t = 5-4x-x^2\), \(\int\frac{dt}{\sqrt{t}} = 2\sqrt{t}\):
\( \displaystyle -\frac{1}{2}\times 2\sqrt{5-4x-x^2} = -\sqrt{5-4x-x^2} \)
For the second part, completing the square:
\(5-4x-x^2 = 9-(x+2)^2\)
\( \displaystyle \int\frac{dx}{\sqrt{9-(x+2)^2}} = \sin^{-1}\left(\frac{x+2}{3}\right) \)
Adding the two parts:
\( \displaystyle I = -\sqrt{5-4x-x^2}+\sin^{-1}\left(\frac{x+2}{3}\right)+C \)
Why the other options are incorrect
Option (2) \(\displaystyle \sqrt{5-4x-x^2}+\sin^{-1}\left(\frac{x+2}{3}\right)+C\) — Incorrect. The multiplier \(-\frac{1}{2}\) makes the root term negative.
Option (3) \(\displaystyle -\frac{1}{2}\sqrt{5-4x-x^2}+\sin^{-1}\left(\frac{x+2}{3}\right)+C\) — Incorrect. It forgets that \(\int\frac{dt}{\sqrt{t}} = 2\sqrt{t}\), so the factor 2 is lost.
Option (4) \(-\sqrt{5-4x-x^2}+C\) — Incorrect. It leaves out the second part, which comes from the constant 1 in the numerator.