Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Integral | Solution: C is an arbitrary constant | ||
| (A) | \(\displaystyle\int\frac{dx}{x^2+25}\) | (I) | \(\displaystyle\frac{1}{10}\log\left|\frac{5+x}{5-x}\right|+C\) |
| (B) | \(\displaystyle\int\frac{dx}{x^2-25}\) | (II) | \(\log|x+\sqrt{x^2-25}|+C\) |
| (C) | \(\displaystyle\int\frac{dx}{25-x^2}\) | (III) | \(\displaystyle\frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right)+C\) |
| (D) | \(\displaystyle\int\frac{dx}{\sqrt{x^{2}-25}}\) | (IV) | \(\displaystyle\frac{1}{10}\log\left|\frac{5-x}{5+x}\right|+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each integral is a standard integral with a = 5.
(A) \(\int\frac{dx}{x^2+25}\)
This matches (III) \(\frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right)+C\):
\(\displaystyle \int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\)
(B) \(\int\frac{dx}{x^2-25}\)
This matches (IV) \(\frac{1}{10}\log\left|\frac{5-x}{5+x}\right|+C\):
\(\displaystyle \int\frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C\)
\(\displaystyle \int\frac{dx}{x^2-25} = \frac{1}{10}\log\left|\frac{x-5}{x+5}\right| + C\)
Since \(|x - 5| = |5 - x|\), this is the same as \(\frac{1}{10}\log\left|\frac{5-x}{5+x}\right| + C\).
(C) \(\int\frac{dx}{25-x^2}\)
This matches (I) \(\frac{1}{10}\log\left|\frac{5+x}{5-x}\right|+C\):
\(\displaystyle \int\frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + C\)
(D) \(\int\frac{dx}{\sqrt{x^{2}-25}}\)
This matches (II) \(\log|x+\sqrt{x^2-25}|+C\):
\(\displaystyle \int\frac{dx}{\sqrt{x^2-a^2}} = \log\left|x + \sqrt{x^2-a^2}\right| + C\)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Tip: (I) and (IV) differ only by a sign, since \(\log\left|\frac{5-x}{5+x}\right| = -\log\left|\frac{5+x}{5-x}\right|\). The sign of x2 decides: \(x^2 - 25\) gives (IV) and \(25 - x^2\) gives (I).
NCERT (2026-27 edition):
Integrals
