Integration of $\int \frac{1}{{x}^{4}+1}\mathrm{d}x$

Blericker74
2022-06-29
Answered

Integration of $\int \frac{1}{{x}^{4}+1}\mathrm{d}x$

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Sophia Mcdowell

Answered 2022-06-30
Author has **14** answers

I think you can do it this way.

$\begin{array}{rl}\int \frac{1}{{x}^{4}+1}\text{}dx& =\frac{1}{2}\cdot \int \frac{2}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot \int \frac{(1-{x}^{2})+(1+{x}^{2})}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot \int \frac{1-{x}^{2}}{1+{x}^{4}}\text{}dx+\frac{1}{2}\int \frac{1+{x}^{2}}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot -\int \frac{1-\frac{1}{{x}^{2}}}{{\textstyle (}x+\frac{1}{x}{)}^{2}-2}\text{}dx+\text{same trick}\end{array}$

$\begin{array}{rl}\int \frac{1}{{x}^{4}+1}\text{}dx& =\frac{1}{2}\cdot \int \frac{2}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot \int \frac{(1-{x}^{2})+(1+{x}^{2})}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot \int \frac{1-{x}^{2}}{1+{x}^{4}}\text{}dx+\frac{1}{2}\int \frac{1+{x}^{2}}{1+{x}^{4}}\text{}dx\\ \text{}& =\frac{1}{2}\cdot -\int \frac{1-\frac{1}{{x}^{2}}}{{\textstyle (}x+\frac{1}{x}{)}^{2}-2}\text{}dx+\text{same trick}\end{array}$

antennense

Answered 2022-07-01
Author has **7** answers

$\int \frac{1}{1+{x}^{4}}dx=\frac{1}{2}\int \frac{1+{x}^{2}+1-{x}^{2}}{1+{x}^{4}}dx$

$\int \frac{1+{x}^{2}}{1+{x}^{4}}dx=\int \frac{\frac{1}{{x}^{2}}+1}{{(x-\frac{1}{x})}^{2}+2}dx$

Set $x-\frac{1}{x}=\sqrt{2}\mathrm{tan}\varphi $

$\int \frac{1-{x}^{2}}{1+{x}^{4}}dx=-\int \frac{1-\frac{1}{{x}^{2}}}{{(x+\frac{1}{x})}^{2}-2}dx$

Set $x+\frac{1}{x}=\sqrt{2}\mathrm{sec}\psi $

$\int \frac{1+{x}^{2}}{1+{x}^{4}}dx=\int \frac{\frac{1}{{x}^{2}}+1}{{(x-\frac{1}{x})}^{2}+2}dx$

Set $x-\frac{1}{x}=\sqrt{2}\mathrm{tan}\varphi $

$\int \frac{1-{x}^{2}}{1+{x}^{4}}dx=-\int \frac{1-\frac{1}{{x}^{2}}}{{(x+\frac{1}{x})}^{2}-2}dx$

Set $x+\frac{1}{x}=\sqrt{2}\mathrm{sec}\psi $

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