What is the integral from-pi to 2x of 1/(1+x^4) ?

The integral from-pi to 2x of 1/(1+x^4) is

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integral from-pi to 2x of 1/(1+x^4)

\int_{- π}^{2 x} \frac{1}{1 + x ^{4}} d x

\frac{1}{2} (2 (\frac{1}{4} ln 8 x^{2} + 42 x + 2 - \frac{1}{4} ln (2 π^{2} - 22 π + 2)) + \frac{1}{2} (arctan (22 x + 1) - arctan (- 2 π + 1))) + \frac{1}{2} (- 2 (\frac{1}{4} ln 8 x^{2} - 42 x + 2 - \frac{1}{4} ln (2 π^{2} + 22 π + 2)) + \frac{1}{2} (arctan (22 x - 1) + arctan (2 π + 1)))

Solution steps

\int_{- π}^{2 x} \frac{1}{1 + x ^{4}} d x

Take the partial fraction of

\frac{1}{1 + x ^{4}} : \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )}

= \int_{- π}^{2 x} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- π}^{2 x} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x + \int_{- π}^{2 x} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

\int_{- π}^{2 x} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{2} (2 (\frac{1}{4} ln 8 x^{2} + 42 x + 2 - \frac{1}{4} ln (2 π^{2} - 22 π + 2)) + \frac{1}{2} (arctan (22 x + 1) - arctan (- 2 π + 1)))

\int_{- π}^{2 x} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{2} (- 2 (\frac{1}{4} ln 8 x^{2} - 42 x + 2 - \frac{1}{4} ln (2 π^{2} + 22 π + 2)) + \frac{1}{2} (arctan (22 x - 1) + arctan (2 π + 1)))

= \frac{1}{2} (2 (\frac{1}{4} ln 8 x^{2} + 42 x + 2 - \frac{1}{4} ln (2 π^{2} - 22 π + 2)) + \frac{1}{2} (arctan (22 x + 1) - arctan (- 2 π + 1))) + \frac{1}{2} (- 2 (\frac{1}{4} ln 8 x^{2} - 42 x + 2 - \frac{1}{4} ln (2 π^{2} + 22 π + 2)) + \frac{1}{2} (arctan (22 x - 1) + arctan (2 π + 1)))

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What is the integral from-pi to 2x of 1/(1+x^4) ?
The integral from-pi to 2x of 1/(1+x^4) is