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 May 12th, 2017, 11:23 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Integral. How do we integrate $\displaystyle \int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$ ? My attempt was changing it to $\displaystyle \int \frac{x^2}{\sqrt{x^4+x^2+1}}dx-\int \frac{1}{x^2\sqrt{x^4+x^2+1}}dx$, but it seems not working.
 May 13th, 2017, 10:30 AM #2 Senior Member   Joined: May 2008 Posts: 259 Thanks: 50 Thanks from limiTS
 May 13th, 2017, 07:35 PM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 I know the answer. But how to solve it?
 May 14th, 2017, 02:15 PM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Anyone?
 May 15th, 2017, 11:44 PM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Anyone?
 May 16th, 2017, 05:07 PM #6 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 Trig substitution works. First, complete the square: $\displaystyle\int\frac{x^4-1}{x^2\sqrt{x^4+2x^2+1-x^2}}dx=\int\frac{x^4-1}{x^2\sqrt{(x^2+1)^2-x^2}}dx$ Factor out and simplify: $\displaystyle\int\frac{x^4-1}{x^2\sqrt{(x^2+1)^2}\sqrt{1-\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{x^4-1}{x^2(x^2+1)\sqrt{1-\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{(x^2+1)(x^2-1)}{x^2(x^2+1)\sqrt{1-\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{\color{red}{x^ 2-1}}{x^2\sqrt{1-\frac{x^2}{(x^2+1)^2}}}\color{red}{dx}$ Let $\displaystyle \color{blue}{\sin\theta}=\frac{\color{blue}{x}}{ \color{blue}{x^2+1}}$. Then $\displaystyle\cos\theta\,d\theta=\frac{x^2+1-2x^2}{(x^2+1)^2}dx=\frac{-x^2+1}{(x^2+1)^2}dx=-\frac{\color{red}{x^2-1}}{(x^2+1)^2}\color{red}{dx}$ Which means that $\displaystyle -(x^2+1)^2\cos\theta\,d\theta=(\color{red}{x^2-1})\color{red}{dx}$ Substitute in the trig: $\displaystyle\int\frac{(x^2-1)}{x^2\sqrt{1-\frac{x^2}{(x^2+1)^2}}}dx\Longrightarrow\int\frac{-(x^2+1)^2\cos\theta}{x^2\sqrt{1-\sin^2\theta}}d\theta=\int-\frac{(x^2+1)^2\cos\theta}{x^2\left|\cos\theta \right|}d\theta$ Assuming that $\cos\theta\gt 0$, the cosines cancel: $\displaystyle\int-\frac{(\color{blue}{x^2+1})^2}{\color{blue}{x}^2}d \theta$, which by another trig substitution is $\displaystyle\int-\frac{1}{\color{blue}{\sin^2\theta}}d\theta$ Then $\displaystyle\int -\frac{1}{\sin^2\theta}d\theta=\int -\csc^2\theta\,d\theta=\cot\theta+C$ Finally, undo the trig substitution to get the final answer in $x$. Thanks from jonah

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