\(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
\(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)
Find \(\displaystyle\int{\frac{{\alpha}x^2-\beta}{{\alpha}^2{x^4}+{\beta}^2}dx}\,, \quad \alpha,\,\beta\in\mathbb{R}^{+}\,.\)
Grigorios Kostakos
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: \(\int\frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx\)
\(\displaystyle{I=\int \frac{\alpha x^2-\beta}{\alpha^2 x^4+\beta^2}dx=}\)
\(\displaystyle{=\int \frac{\alpha x^2-\beta}{\left(\alpha x^2+\beta\right)^2-2\alpha\beta x^2}dx}\)
\(\displaystyle{=\int \displaystyle{\frac{\alpha-\frac{\beta}{x^2}}{\left(\alpha x+\frac{\beta}{x}\right)^2-2\alpha\beta}}dx}\).
By the substitution,
\(\displaystyle{\alpha x+\frac{\beta}{x}=t}\),we have \(\displaystyle{dt=\left(\alpha-\frac{\beta}{x^2}\right)dx}\)
So,
\(\displaystyle{I=\int \frac{1}{t^2-2\alpha\beta}dt=}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\int \frac{(t+\sqrt{2\alpha\beta})-(t-\sqrt{2\alpha\beta})}{(t-\sqrt{2\alpha\beta})(t+\sqrt{\alpha\beta})}dt}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\left[\int \frac{1}{t-\sqrt{2\alpha\beta}}dt-\int \frac{1}{t+\sqrt{2\alpha\beta}}dt\right]}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{t-\sqrt{2\alpha\beta}}{t+\sqrt{2\alpha\beta}}\right)+c}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{\alpha x^2-\sqrt{2\alpha\beta}x+\beta}{\alpha x^2+\sqrt{2\alpha\beta}x+\beta}\right)+c,c\in\mathbb{R}}\)
\(\displaystyle{=\int \frac{\alpha x^2-\beta}{\left(\alpha x^2+\beta\right)^2-2\alpha\beta x^2}dx}\)
\(\displaystyle{=\int \displaystyle{\frac{\alpha-\frac{\beta}{x^2}}{\left(\alpha x+\frac{\beta}{x}\right)^2-2\alpha\beta}}dx}\).
By the substitution,
\(\displaystyle{\alpha x+\frac{\beta}{x}=t}\),we have \(\displaystyle{dt=\left(\alpha-\frac{\beta}{x^2}\right)dx}\)
So,
\(\displaystyle{I=\int \frac{1}{t^2-2\alpha\beta}dt=}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\int \frac{(t+\sqrt{2\alpha\beta})-(t-\sqrt{2\alpha\beta})}{(t-\sqrt{2\alpha\beta})(t+\sqrt{\alpha\beta})}dt}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\left[\int \frac{1}{t-\sqrt{2\alpha\beta}}dt-\int \frac{1}{t+\sqrt{2\alpha\beta}}dt\right]}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{t-\sqrt{2\alpha\beta}}{t+\sqrt{2\alpha\beta}}\right)+c}\)
\(\displaystyle{=\frac{1}{2\sqrt{2\alpha\beta}}\ln \left(\frac{\alpha x^2-\sqrt{2\alpha\beta}x+\beta}{\alpha x^2+\sqrt{2\alpha\beta}x+\beta}\right)+c,c\in\mathbb{R}}\)
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 7 guests