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Theorems on Integration

Evaluate: ∫x2...

Question

Evaluate: $\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} dx$

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Solution

$\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} dx$
Substituting $x^{2} = t$ for partial fraction,
$\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{t}{(t + a^{2}) (t + b^{2})}$
Let $\frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{(t + a^{2})} + \frac{B}{(t + b^{2})}$
$\Rightarrow t = A (t + b^{2}) + B (t + a^{2})$
Substituting $t = - a^{2}$ ,
$\Rightarrow - a^{2} = A (- a^{2} + b^{2})$
$\Rightarrow A = \frac{a^{2}}{(a^{2} - b^{2})}$
Substituting $t = - b^{2}$ ,
$\Rightarrow - b^{2} = B (- b^{2} + a^{2})$
$\Rightarrow B = - \frac{b^{2}}{(a^{2} - b^{2})}$
$∴ \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} dx$
$= \frac{a^{2}}{(a^{2} - b^{2})} \int \frac{1}{(x^{2} + a^{2})} dx - \frac{b^{2}}{(a^{2} - b^{2})} \int \frac{1}{(x^{2} + b^{2})} dx$
$= \frac{a^{2}}{a^{2} - b^{2}} [\frac{1}{a} {tan}^{- 1} (\frac{x}{a})] - \frac{b^{2}}{a^{2} - b^{2}} [\frac{1}{b} {tan}^{- 1} (\frac{x}{b})]$
$[∴ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} {tan}^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{(a^{2} - b^{2})} [a {tan}^{- 1} (\frac{x}{a}) - b {tan}^{- 1} (\frac{x}{b})] + c$
where, c is constant of integration
Hence, the integration is
$\frac{1}{(a^{2} - b^{2})} [a {tan}^{- 1} (\frac{x}{a}) - b {tan}^{- 1} (\frac{x}{b})] + c$

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Theorems on Integration

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