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- Jan 26, 2012

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Problem: find

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

Answer: It's perhaps a not-so-well-known fact that, although $x^{2}+y^{2}$ does not factor over the reals, $x^{4}+y^{4}$ does. In fact,

$$x^{4}+y^{4}= \left(x^{2}+ \sqrt{2} \, xy+y^{2} \right) \left(x^{2}- \sqrt{2} \, xy+y^{2} \right).$$

Hence, we have

$$\int \frac{dx}{1+x^{4}}=\int \frac{dx}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}.$$

Next, you can use partial fractions to pull them apart. That is, assume that you can write

$$\frac{1}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}

= \frac{Ax+B}{x^{2}+ \sqrt{2} \, x+1}+ \frac{Cx+D}{x^{2}- \sqrt{2} \, x+1}.$$

Once you've done all that algebra, you wind up with

$$ \frac{1}{1+x^{4}}=

\frac{ (\sqrt{2} \,x+2)/4}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (\sqrt{2} \,x-2)/4}{x^{2}- \sqrt{2} \, x+1}=\frac{ (2x+2 \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x-2 \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}.$$

The final trick is as follows: you write each of these two fractions as two more fractions (so you have four total); one of each of them you write so that the derivative of the denominator shows up in the numerator (setting up a $\ln$-type integral), and then the other one has just a constant in the numerator - so you can complete the square and get an $\tan^{-1}$-type integral. You'd get this:

$$ \frac{1}{1+x^{4}}=

\frac{ (2x+ \sqrt{2}+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x- \sqrt{2}- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{ (\sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

-\frac{ (- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{1/4}{(x^{2}+ \sqrt{2} \, x+1/2)+1/2}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

+\frac{ 1/4}{(x^{2}- \sqrt{2} \, x+1/2)+1/2}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{1/4}{(x+ \sqrt{2}/2)^{2}+1/2}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

+\frac{ 1/4}{(x- \sqrt{2}/2)^{2}+1/2}.$$

This succumbs to standard $u$-substitutions and knowledge of the $\tan^{-1}$ derivative.

The final answer is

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

$$=\frac{ \sqrt{2}}{8} \ln \left(x^{2}+ \sqrt{2} \, x + 1 \right)

+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x+1 \right)

-\frac{ \sqrt{2}}{8} \ln \left(x^{2}- \sqrt{2} \, x + 1 \right)

+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x-1 \right)+C.$$

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-antiderivative-1-1-x^4-4207.html

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

Answer: It's perhaps a not-so-well-known fact that, although $x^{2}+y^{2}$ does not factor over the reals, $x^{4}+y^{4}$ does. In fact,

$$x^{4}+y^{4}= \left(x^{2}+ \sqrt{2} \, xy+y^{2} \right) \left(x^{2}- \sqrt{2} \, xy+y^{2} \right).$$

Hence, we have

$$\int \frac{dx}{1+x^{4}}=\int \frac{dx}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}.$$

Next, you can use partial fractions to pull them apart. That is, assume that you can write

$$\frac{1}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}

= \frac{Ax+B}{x^{2}+ \sqrt{2} \, x+1}+ \frac{Cx+D}{x^{2}- \sqrt{2} \, x+1}.$$

Once you've done all that algebra, you wind up with

$$ \frac{1}{1+x^{4}}=

\frac{ (\sqrt{2} \,x+2)/4}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (\sqrt{2} \,x-2)/4}{x^{2}- \sqrt{2} \, x+1}=\frac{ (2x+2 \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x-2 \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}.$$

The final trick is as follows: you write each of these two fractions as two more fractions (so you have four total); one of each of them you write so that the derivative of the denominator shows up in the numerator (setting up a $\ln$-type integral), and then the other one has just a constant in the numerator - so you can complete the square and get an $\tan^{-1}$-type integral. You'd get this:

$$ \frac{1}{1+x^{4}}=

\frac{ (2x+ \sqrt{2}+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x- \sqrt{2}- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{ (\sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

-\frac{ (- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{1/4}{(x^{2}+ \sqrt{2} \, x+1/2)+1/2}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

+\frac{ 1/4}{(x^{2}- \sqrt{2} \, x+1/2)+1/2}$$

$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}

+ \frac{1/4}{(x+ \sqrt{2}/2)^{2}+1/2}

- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}

+\frac{ 1/4}{(x- \sqrt{2}/2)^{2}+1/2}.$$

This succumbs to standard $u$-substitutions and knowledge of the $\tan^{-1}$ derivative.

The final answer is

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

$$=\frac{ \sqrt{2}}{8} \ln \left(x^{2}+ \sqrt{2} \, x + 1 \right)

+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x+1 \right)

-\frac{ \sqrt{2}}{8} \ln \left(x^{2}- \sqrt{2} \, x + 1 \right)

+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x-1 \right)+C.$$

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-antiderivative-1-1-x^4-4207.html

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