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Match List-I with List-II

List-IList-II
Indefinite integralSolution (where 'c' is an arbitrary constant)
(A) ∫x2−a2 dx\displaystyle\int \sqrt{x^2-a^2} \, dx(I) 1atan⁡−1xa+c\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c
(B) ∫1x2−a2 dx\displaystyle\int \dfrac{1}{x^2-a^2} \, dx(II) x2x2+a2+a22log⁡[x+x2+a2]+c\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c
(C) ∫1x2+a2 dx\displaystyle\int \dfrac{1}{x^2+a^2} \, dx(III) x2x2−a2−a22log⁡[x+x2−a2]+c\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c
(D) ∫x2+a2 dx\displaystyle\int \sqrt{x^2+a^2} \, dx(IV) 12alog⁡∣x−ax+a∣+C\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

These are standard forms. The radical x2−a2\sqrt{x^2-a^2} integrates with a minus sign before the log term, matching (III); 1x2−a2\dfrac{1}{x^2-a^2} gives the log of a ratio, matching (IV); 1x2+a2\dfrac{1}{x^2+a^2} gives the arctangent form (I); and x2+a2\sqrt{x^2+a^2} gives (II).

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