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Question Number 114161 by bemath last updated on 17/Sep/20

$$\int \frac{dx}{x^4-5x^2-16}$$

Answered by Olaf last updated on 17/Sep/20

$$-\frac{2}{\sqrt{89}}\frac{\arctan \left(\frac{\sqrt{2}x}{\sqrt{\sqrt{89}-5}}\right)}{\sqrt{\sqrt{89}-5}}-\frac{2}{\sqrt{89}}\frac{\arctan \left(\frac{\sqrt{2}x}{\sqrt{\sqrt{89}+5}}\right)}{\sqrt{\sqrt{89}+5}}$$

Answered by malwan last updated on 17/Sep/20

$$\int \frac{dx}{[x^2-\left(\frac{5+\sqrt{89}}{2}\right)][x^2-\left(\frac{5-\sqrt{89}}{2}\right)]}$$$$=\int \frac{dx}{(x+\sqrt{\frac{5+\sqrt{89}}{2}})(x-\sqrt{\frac{5+\sqrt{89}}{2}})(x+\sqrt{\frac{5-\sqrt{89}}{2}})(x-\sqrt{\frac{5-\sqrt{89}}{2}})}$$$$=\int \frac{dx}{(x+a)(x-a)(x+b)(x-b)}$$$$=\frac{1}{2a(a+b)(-a+b)}ln\mid x+a\mid +$$$$\frac{1}{2a(a+b)(a-b)}ln\mid x-a\mid +$$$$\frac{1}{2b(a+b)(a-b)}ln\mid x+b\mid +$$$$\frac{1}{2b(-a+b)(a+b)}ln\mid x-b\mid$$$$=\frac{1}{2a(b^2-a^2)}ln\mid x+a\mid +$$$$\frac{1}{2a(a^2-b^2)}ln\mid x-a\mid +$$$$\frac{1}{2b(a^2-b^2)}ln\mid x+b\mid +$$$$\frac{1}{2b(b^2-a^2)}ln\mid x-b\mid$$$$nowa=\sqrt{\frac{5+\sqrt{89}}{2}};b=\sqrt{\frac{5-\sqrt{89}}{2}}$$$$a^2-b^2=\sqrt{89};b^2-a^2=-\sqrt{89}$$

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