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Miscellaneous Exercise 4 · Q73

Q.Evaluate: ∫0a1a2+ax−x2 dx\int_0^a \dfrac{1}{a^2+ax-x^2}\,dx

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a2+ax−x2=−[x2−ax−a2]=5a24−(x−a2)2a^2+ax-x^2=-\big[x^2-ax-a^2\big]=\dfrac{5a^2}4-\Big(x-\dfrac a2\Big)^2.

∫0adx5a24−(x−a2)2=[15 aln⁡∣5a2+x−a25a2−x+a2∣]0a.\int_0^a\frac{dx}{\frac{5a^2}4-(x-\frac a2)^2}=\Big[\frac1{\sqrt5\,a}\ln\Big|\frac{\frac{\sqrt5a}2+x-\frac a2}{\frac{\sqrt5a}2-x+\frac a2}\Big|\Big]_0^a.

At x=ax=a: ratio =a(5+1)/2a(5−1)/2=5+15−1=\dfrac{a(\sqrt5+1)/2}{a(\sqrt5-1)/2}=\dfrac{\sqrt5+1}{\sqrt5-1}. At x=0x=0: ratio =5−15+1=\dfrac{\sqrt5-1}{\sqrt5+1} (the reciprocal). …

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