The problem is the integral of

$\int \frac{-8x}{{x}^{4}-{a}^{4}}\phantom{\rule{thinmathspace}{0ex}}dx$

$\int \frac{-8x}{{x}^{4}-{a}^{4}}\phantom{\rule{thinmathspace}{0ex}}dx$

juanberrio8a
2022-06-21
Answered

The problem is the integral of

$\int \frac{-8x}{{x}^{4}-{a}^{4}}\phantom{\rule{thinmathspace}{0ex}}dx$

$\int \frac{-8x}{{x}^{4}-{a}^{4}}\phantom{\rule{thinmathspace}{0ex}}dx$

You can still ask an expert for help

humusen6p

Answered 2022-06-22
Author has **22** answers

Noting that

$I=\int \frac{-8x}{{x}^{4}-{a}^{4}}dx=-4\int \frac{d\left({x}^{2}\right)}{{\left({x}^{2}\right)}^{2}-{\left({a}^{2}\right)}^{2}}$

Let $y={x}^{2}$ and $b={a}^{2}$, then

$\begin{array}{rl}I& =-4\int \frac{dy}{{y}^{2}-{b}^{2}}\\ & =-4\int \frac{1}{2b}(\frac{1}{y-b}-\frac{1}{y+b})dy\\ & =-\frac{2}{b}(\mathrm{ln}|y-b|-\mathrm{ln}|y+b|)+C\\ & =-\frac{2}{{a}^{2}}\mathrm{ln}\left|\frac{{x}^{2}-{a}^{2}}{{x}^{2}+{a}^{2}}\right|+C\\ & =\frac{2}{{a}^{2}}\mathrm{ln}\left|\frac{{x}^{2}+{a}^{2}}{{x}^{2}-{a}^{2}}\right|+C\end{array}$

$I=\int \frac{-8x}{{x}^{4}-{a}^{4}}dx=-4\int \frac{d\left({x}^{2}\right)}{{\left({x}^{2}\right)}^{2}-{\left({a}^{2}\right)}^{2}}$

Let $y={x}^{2}$ and $b={a}^{2}$, then

$\begin{array}{rl}I& =-4\int \frac{dy}{{y}^{2}-{b}^{2}}\\ & =-4\int \frac{1}{2b}(\frac{1}{y-b}-\frac{1}{y+b})dy\\ & =-\frac{2}{b}(\mathrm{ln}|y-b|-\mathrm{ln}|y+b|)+C\\ & =-\frac{2}{{a}^{2}}\mathrm{ln}\left|\frac{{x}^{2}-{a}^{2}}{{x}^{2}+{a}^{2}}\right|+C\\ & =\frac{2}{{a}^{2}}\mathrm{ln}\left|\frac{{x}^{2}+{a}^{2}}{{x}^{2}-{a}^{2}}\right|+C\end{array}$

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Can the Risch-algorithm actually prove that ${e}^{-{x}^{2}}$ has no closed-form antiderivative?

The Risch algorithm is used to find closed-form antiderivatives.

If I understand the article right, only heuristics are known. On the other hand, I came across the claim that it is known that ${e}^{-{x}^{2}}$ has no closed-form antiderivative.

Is the Risch-algorithm successful in the case $f(x)={e}^{-{x}^{2}}$? Is it actually proven that no closed-form antiderivative exist or just very probable because no form has been found using the Risch algorithm ?

This question could well be a duplicate, but I am not sure whether the aspect of decidability has been asked.

The Risch algorithm is used to find closed-form antiderivatives.

If I understand the article right, only heuristics are known. On the other hand, I came across the claim that it is known that ${e}^{-{x}^{2}}$ has no closed-form antiderivative.

Is the Risch-algorithm successful in the case $f(x)={e}^{-{x}^{2}}$? Is it actually proven that no closed-form antiderivative exist or just very probable because no form has been found using the Risch algorithm ?

This question could well be a duplicate, but I am not sure whether the aspect of decidability has been asked.