\(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)

#1

Post by Grigorios Kostakos »

Find \(\displaystyle\int{\frac{{\alpha}x^2-\beta}{{\alpha}^2{x^4}+{\beta}^2}dx}\,, \quad \alpha,\,\beta\in\mathbb{R}^{+}\,.\)
Grigorios Kostakos
Papapetros Vaggelis
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Re: \(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)

#2

Post by Papapetros Vaggelis »

\(\displaystyle{I=\int \frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx=}\)

\(\displaystyle{=\int \frac{\alpha x^2-\beta}{\left(\alpha x^2+\beta\right)^2-2\alpha\beta x^2}dx}\)

\(\displaystyle{=\int \displaystyle{\frac{\alpha-\frac{\beta}{x^2}}{\left(\alpha x+\frac{\beta}{x}\right)^2-2\alpha\beta}}dx}\).

By the substitution,

\(\displaystyle{\alpha x+\frac{\beta}{x}=t}\),we have \(\displaystyle{dt=\left(\alpha-\frac{\beta}{x^2}\right)dx}\)

So,

\(\displaystyle{I=\int \frac{1}{t^2-2\alpha\beta}dt=}\)

\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\int \frac{(t+\sqrt{2\alpha\beta})-(t-\sqrt{2\alpha\beta})}{(t-\sqrt{2\alpha\beta})(t+\sqrt{\alpha\beta})}dt}\)

\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\left[\int \frac{1}{t-\sqrt{2\alpha\beta}}dt-\int \frac{1}{t+\sqrt{2\alpha\beta}}dt\right]}\)

\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{t-\sqrt{2\alpha\beta}}{t+\sqrt{2\alpha\beta}}\right)+c}\)

\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{\alpha x^2-\sqrt{2\alpha\beta}x+\beta}{\alpha x^2+\sqrt{2\alpha\beta}x+\beta}\right)+c,c\in\mathbb{R}}\)
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