How to compute $\int_0^1 \frac{1}{1+x^4}\;dx$? I am having problems trying to compute $$\int_0^1 \frac{1}{1+x^4}\;dx$$
Wolfram alpha gives an answer $$\frac{\pi + 2 \coth^{-1}(\sqrt{2})}{4 \sqrt{2}}$$
 A: Factor as $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt 2 \cdot x)^2=(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x +1)$ and use partial fractions.
A: Hint: write the numerator as $\dfrac{1+x^2+1-x^2}2$
Divide both numerator & the denominator by $x^2$
A: Notice, $$\int_{0}^{1}\frac{dx}{1+x^4}$$
$$=\int_{0}^{1}\frac{\frac{1}{x^2}dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\frac{2}{x^2}dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}\right]$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+2}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2-2}\right]$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+(\sqrt{2})^2}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2-(\sqrt{2})^2}\right]$$
$$=\frac{1}{2}\left[\frac{1}{\sqrt 2}\left\{\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt 2}\right)\right\}_{0}^{1}-\left\{\frac{1}{2\sqrt 2}\ln\left|\frac{x+\frac{1}{x}-\sqrt 2}{x+\frac{1}{x}+\sqrt 2}\right|\right\}_{0}^{1}\right]$$
$$=\frac{1}{2}\left[\frac{1}{\sqrt 2}\left\{\tan^{-1}\left(0\right)-\tan^{-1}\left(\infty\right)\right\}-\frac{1}{2\sqrt 2}\left\{\ln\left|\frac{2-\sqrt 2}{2+\sqrt 2}\right|-\ln|1|\right\}\right]$$
$$=\frac{1}{2}\left[-\frac{\pi}{2\sqrt 2}-\frac{1}{\sqrt 2}\ln\left|\frac{2-\sqrt 2}{2}\right|\right]$$
$$=-\frac{\pi}{4\sqrt 2}-\frac{1}{2\sqrt 2}\ln\left|\frac{2-\sqrt 2}{2}\right|$$
