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May 12th, 2017, 11:23 PM   #1
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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.
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May 13th, 2017, 10:30 AM   #2
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May 13th, 2017, 07:35 PM   #3
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I know the answer. But how to solve it?
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May 14th, 2017, 02:15 PM   #4
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Anyone?
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May 15th, 2017, 11:44 PM   #5
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May 16th, 2017, 05:07 PM   #6
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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 $x$.
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