Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{dx}{x^2+2x+5}\) | (I) | \(\displaystyle\frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C\) |
| (B) | \(\int\displaystyle\frac{dx}{\sqrt{x^2+2x+5}}\) | (II) | \(\displaystyle\frac{1}{4}\log\left|\frac{x-1}{x+3}\right|+C\) |
| (C) | \(\int\displaystyle\frac{dx}{\sqrt{3-2x-x^2}}\) | (III) | \(\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\) |
| (D) | \(\int\displaystyle\frac{dx}{x^2+2x-3}\) | (IV) | \(\displaystyle\sin^{-1}\left(\frac{x+1}{2}\right)+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
Each quadratic becomes a perfect square in \(t = x+1\) plus or minus 4; then the matching standard integral with a = 2 is used.
(A) \(\int\frac{dx}{x^2+2x+5}\)
This matches (I) \(\frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C\).
\(x^2+2x+5 = (x+1)^2+4\)
\( \displaystyle \int\frac{dt}{t^2+2^2} = \frac{1}{2}\tan^{-1}\frac{t}{2}+C \)
(B) \(\int\frac{dx}{\sqrt{x^2+2x+5}}\)
This matches (III) \(\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\).
\( \displaystyle \int\frac{dt}{\sqrt{t^2+2^2}} = \log\left|t+\sqrt{t^2+4}\right|+C \)
With \(t = x+1\), \(t^2+4 = x^2+2x+5\).
(C) \(\int\frac{dx}{\sqrt{3-2x-x^2}}\)
This matches (IV) \(\sin^{-1}\left(\frac{x+1}{2}\right)+C\).
\(3-2x-x^2 = 4-(x+1)^2\)
\( \displaystyle \int\frac{dt}{\sqrt{2^2-t^2}} = \sin^{-1}\frac{t}{2}+C \)
(D) \(\int\frac{dx}{x^2+2x-3}\)
This matches (II) \(\frac{1}{4}\log\left|\frac{x-1}{x+3}\right|+C\).
\(x^2+2x-3 = (x+1)^2-4\)
\( \displaystyle \int\frac{dt}{t^2-2^2} = \frac{1}{4}\log\left|\frac{t-2}{t+2}\right|+C \)
With \(t = x+1\), \(t-2 = x-1\) and \(t+2 = x+3\).