\(\int\displaystyle\frac{x^2+4}{(x^2+3)(x^2+5)}\,dx=u\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+v\tan^{-1}\left(\frac{x}{\sqrt{5}}\right)+c\), where c is arbitrary constant, then value of \(\displaystyle\frac{1}{u^2}+\frac{1}{v^2}\) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 32
The integrand contains x only through \(x^2\), so for the partial fractions put \(x^2 = t\):
\( \displaystyle \frac{t+4}{(t+3)(t+5)} = \frac{A}{t+3} + \frac{B}{t+5} \)
\(t + 4 = A(t+5) + B(t+3)\)
\( \displaystyle t = -3: \ 1 = 2A, \quad A = \frac{1}{2} \)
\( \displaystyle t = -5: \ -1 = -2B, \quad B = \frac{1}{2} \)
So
\( \displaystyle \int\frac{x^2+4}{(x^2+3)(x^2+5)}\,dx = \frac{1}{2}\int\frac{dx}{x^2+3} + \frac{1}{2}\int\frac{dx}{x^2+5} \)
Using \(\displaystyle\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}\):
\( \displaystyle = \frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c \)
Comparing:
\( \displaystyle u = \frac{1}{2\sqrt{3}}, \quad v = \frac{1}{2\sqrt{5}} \)
\( \displaystyle \frac{1}{u^2} + \frac{1}{v^2} = 12 + 20 = 32 \)