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Methods of integration

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Question

$If \int \frac{x^{2} - 4}{x^{4} + 9 x^{2} + 16} d x = A \tan^{- 1} (f (x)) + B, then the value of f (10) is$

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Solution

$I = \int \frac{x^{2} - 4}{x^{4} + 9 x^{2} + 16} d x$
Divide both numerator and denominator by x²
$\begin{array}{l} I = \int \frac{(1 - \frac{4}{x^{2}})}{x^{2} + 9 + \frac{16}{x^{2}}} dx = \int \frac{(1 - \frac{4}{x^{2}})}{{(x + \frac{4}{x})}^{2} + 1} dx \\ Put u = x + \frac{4}{x}, du = (1 - \frac{4}{x^{2}}) dx \\ I = \int \frac{du}{u^{2} + 1} = \tan^{- 1} (u) + B \\ = 1 \cdot \tan^{- 1} (x + \frac{4}{x}) + B \\ \Rightarrow f (x) = x + \frac{4}{x} \end{array}$

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