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If C is an arbitrary constant, then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.∫x2+a2 dx\displaystyle\int \sqrt{x^2+a^2}\, dxI.x2a2−x2+a22sin⁡−1xa+C\dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C
B.∫a2−x2 dx\displaystyle\int \sqrt{a^2-x^2}\, dxII.log⁡∣x+x2+a2∣+C\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C
C.∫1x2+a2 dx\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\, dxIII.sin⁡−1xa+C\sin^{-1}\dfrac{x}{a} + C
D.∫1a2−x2 dx\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\, dxIV.x2x2+a2+a22log⁡∣x+x2+a2∣+C\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

These are the four standard results. The radical integrals produce a half-product term plus a logarithm or inverse sine, so A matches IV and B matches I. The reciprocal radicals give log⁡∣x+x2+a2∣+C\log\left|x+\sqrt{x^2+a^2}\right|+C for C (II) and sin⁡−1xa+C\sin^{-1}\frac{x}{a}+C for D (III).

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