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Addition of Rational Expressions

∫ 0 ∞ d t /...

Question

$\int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + 4 t^{2})}$

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Solution

$\begin{matrix} \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + 4 t^{2})} p u t t^{2} = y \Rightarrow \frac{1}{(1 + y) (1 + 4 y)} = \frac{A}{(1 + y)} + \frac{B}{(1 + 4 y)} \Rightarrow \frac{1}{(1 + y) (1 + 4 y)} = \frac{A + 4 A Y + B + B Y}{(1 + y) (1 + 4 y)} A + B = 1 - 4 A = B P u t B = - 4 A i n e q u a t i o n (i) \Rightarrow A - 4 A = 1 A = \frac{- 1}{3} a n d B = \frac{4}{3} \Rightarrow \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + 4 t^{2})} = \frac{- 1}{3} \int_{0}^{\infty} \frac{1}{1 + t^{2}} d t + \frac{4}{3} \int_{0}^{\infty} \frac{d t}{1 + 4 t^{2}} \Rightarrow \frac{- 1}{3} \int_{0}^{\infty} \frac{1}{(1 + t^{2})} d t + \frac{1}{3} \int_{0}^{\infty} \frac{d t}{[{(\frac{1}{2})}^{2} + t^{2}]} = [\frac{- 1}{3} {tan}^{- 1} t + \frac{1}{3} \times 2 {tan}^{- 1} 2 t] + c \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + 4 t^{2}) = \frac{1}{3}} [2 {tan}^{- 1} 2 t - {tan}^{- 1}] + c = \frac{1}{3} [2 {tan}^{- 1} \infty - {tan}^{- 1} \infty - 0] + c = \frac{1}{3} [π - \frac{π}{2} - 0] = \frac{π}{6} \end{matrix}$

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