Skip to main contentSkip to solution

Match List-I with List-II

Where C is arbitrary constant.

List-IList-II
(A) ∫dxx2−a2=\displaystyle\int \dfrac{dx}{x^2 - a^2} =(I) log⁡∣x+x2+a2∣+C\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C
(B) ∫dxa2−x2=\displaystyle\int \dfrac{dx}{a^2 - x^2} =(II) log⁡∣x+x2−a2∣+C\log\left\vert x + \sqrt{x^2-a^2}\right\vert + C
(C) ∫dxx2−a2=\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} =(III) 12alog⁡∣x−ax+a∣+C\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a}\right\vert + C
(D) ∫dxx2+a2=\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} =(IV) 12alog⁡∣a+xa−x∣+C\dfrac{1}{2a}\log\left\vert \dfrac{a+x}{a-x}\right\vert + C

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

These are the standard forms.

∫dxx2−a2=12alog⁡∣x−ax+a∣+C\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C gives (A)-(III), and ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C gives (B)-(IV).

The surd forms give (C)-(II) and (D)-(I).

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question