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Exercise 4.2 · Q15

Q.Evaluate: ∫−421x2+4x+13 dx\int_{-4}^{2} \dfrac{1}{x^2+4x+13}\,dx

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∫−42dx(x+2)2+9=[13tan⁡−1x+23]−42=13[tan⁡−143−tan⁡−1( ⁣−23)]=13[tan⁡−143+tan⁡−123].\int_{-4}^2\frac{dx}{(x+2)^2+9}=\Big[\frac13\tan^{-1}\frac{x+2}3\Big]_{-4}^2=\frac13\Big[\tan^{-1}\frac43-\tan^{-1}\Big(\!-\frac23\Big)\Big]=\frac13\Big[\tan^{-1}\frac43+\tan^{-1}\frac23\Big]. …

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