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I've seen other methods for integrating this on StackExchange yet I'm wondering if what I'm doing is valid as well.

$$\int \frac {1}{x^4 + 1} dx$$

$$\int \frac {1}{(x^2 + 1)(x^2-1)+2} dx$$

Using the trigonometric substitution: $x = \tan(u), \,dx = \sec^2(u)\, du$

$$\int \frac {\sec^2u}{((\tan^2(u) +1)(\tan^2(u)-1)+2) } du$$

Using trigonometric identities: $ \tan^2(u) +1= \sec^2(u)$, and completing the trigonometric substitution the integral becomes:

edit: removed brackets around variable for readability

$$\int \frac {\sec^2u}{(\,(\sec^2u)\ ((\,(\sec^2u-1)-1)+2) } du$$

$$\int \frac {\sec^2u}{(\,(\sec^2u\,)(\sec^2u-2)+2) } du$$

Proceeding into the unknown:

$$\int \frac {\sec^2u}{((\sec^4u-2\sec^2u)+2) } du$$

$$\int \frac {\sec^2u}{(\sec^4u-2\sec^2u + 2) } du$$

Making a substitution $ v=\sec^2u$ would not simplify this any further as far as I can see since: $ dv = 2(\sec^2u)( \tan u)\, du$


Previous working:

$$\int \cos^2(u)\, du$$

$$\frac{1}{4}(2u+\sin(2u))$$

Using $\sin(2u) = 2\sin(u)\cos(u)$ & back substitution of: $u = \arctan(x)$

$$\frac{1}{4}(2\arctan(x)) * 2\sin(\arctan(x))\cos(\arctan(x))$$

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    $\begingroup$ You're made an error in cancelling the $\sec^2(u)$. You may also want to check that the parentheses are matched correctly on the step before that. $\endgroup$ – Irregular User Mar 7 '16 at 21:16
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    $\begingroup$ Look at here and here $\endgroup$ – user249332 Mar 7 '16 at 21:21
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    $\begingroup$ Even with the new edit (the bracketing is better) the $sec^2(u)$ has been cancelled incorrectly. What happened to the $2$? This is like saying that $\frac{a}{a+b} = \frac{1}{1 + b}$ $\endgroup$ – Irregular User Mar 7 '16 at 21:31
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    $\begingroup$ You're welcome! It just goes to show the importance of keeping good track of your work, whether that be through the use of potentially different bracketing (e.g. { } ( ) [ ] or whatever you wish to invent) or not working purely in LaTeX (write on paper first!). Also no, some of the people who answered chose to delete their own answer so you editing your post did nothing to those. $\endgroup$ – Irregular User Mar 7 '16 at 22:14
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    $\begingroup$ Possible duplicate of Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ $\endgroup$ – Nosrati Oct 3 '18 at 5:52
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NOTE:

The solution herein addressed the Originally Posted Question the OP edited the question.

Your mistake starts at the third line,

$$\int \frac {1}{x^4 + 1} dx$$

$$\int \frac {1}{(x^2 + 1)(x^2-1)+2} dx$$

Using the trigonometric substitution: $x = \tan(u), \,dx = \sec^2(u)\, du$

$$\int \frac {\color{red}{\sec^2 u }}{((\tan^2(u) +1)(\tan^2(u)-1)+2) } \color{red}{du}$$

See more at the duplicate

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    $\begingroup$ You may want to recheck his fifth line more closely. $\endgroup$ – Irregular User Mar 7 '16 at 21:22
  • $\begingroup$ Ah right, let me fix that. $\endgroup$ – mostlyfabulous Mar 7 '16 at 22:01

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