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

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

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

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

\frac{ln ( \frac{6 - 2}{2} ) + 2 ( arctan ( 2 2 + 1 ) - arctan ( 2 + 1 ) ) - ln ( \frac{6 + 2}{2} ) + 2 ( arctan ( 2 2 - 1 ) - arctan ( 2 - 1 ) )}{4 2}

Decimal

0.20315 \dots

Solution steps

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

\int_{1}^{2} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{2} (\frac{1}{2 2} ln (\frac{6 - 2}{2}) + \frac{1}{2} (arctan (22 + 1) - arctan (2 + 1)))

\int_{1}^{2} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{2} (- \frac{1}{2 2} ln (\frac{6 + 2}{2}) + \frac{1}{2} (arctan (22 - 1) - arctan (2 - 1)))

= \frac{1}{2} (\frac{1}{2 2} ln (\frac{6 - 2}{2}) + \frac{1}{2} (arctan (22 + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} ln (\frac{6 + 2}{2}) + \frac{1}{2} (arctan (22 - 1) - arctan (2 - 1)))

Simplify

\frac{1}{2} (\frac{1}{2 2} ln (\frac{6 - 2}{2}) + \frac{1}{2} (arctan (22 + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} ln (\frac{6 + 2}{2}) + \frac{1}{2} (arctan (22 - 1) - arctan (2 - 1))) : \frac{ln ( \frac{6 - 2}{2} ) + 2 ( arctan ( 2 2 + 1 ) - arctan ( 2 + 1 ) ) - ln ( \frac{6 + 2}{2} ) + 2 ( arctan ( 2 2 - 1 ) - arctan ( 2 - 1 ) )}{4 2}

= \frac{ln ( \frac{6 - 2}{2} ) + 2 ( arctan ( 2 2 + 1 ) - arctan ( 2 + 1 ) ) - ln ( \frac{6 + 2}{2} ) + 2 ( arctan ( 2 2 - 1 ) - arctan ( 2 - 1 ) )}{4 2}

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