Skip to main contentSkip to solution

∫x2+4(x2+3)(x2+5) dx=utan⁡−1(x3)+vtan⁡−1(x5)+c\int \frac{x^2 + 4}{(x^2 + 3)(x^2 + 5)} \, dx = u \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + v \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c, where c is arbitrary constant, then value of 1u2+1v2\frac{1}{u^2} + \frac{1}{v^2} is equal to :

Solution

✅ Correct Option: 1

Partial fractions: x2+4(x2+3)(x2+5)=Ax2+3+Bx2+5\frac{x^2+4}{(x^2+3)(x^2+5)} = \frac{A}{x^2+3} + \frac{B}{x^2+5}.

Solving: A+B=1A + B = 1 and 5A+3B=45A + 3B = 4, giving A=B=1/2A = B = 1/2.

Integrating: u=123u = \frac{1}{2\sqrt{3}}, v=125v = \frac{1}{2\sqrt{5}}.

1u2+1v2=12+20=32\frac{1}{u^2} + \frac{1}{v^2} = 12 + 20 = 32.

Keyboard Shortcuts

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