What is the integral from 0 to 1 of (1+x^2)/(1+x^4) ?

The integral from 0 to 1 of (1+x^2)/(1+x^4) is (8arctan(sqrt(2)+1)+8arctan(sqrt(2)-1))/(8sqrt(2))

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integral from 0 to 1 of (1+x^2)/(1+x^4)

\int_{0}^{1} \frac{1 + x ^{2}}{1 + x ^{4}} d x

\frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

Decimal Notation

1.11072 \dots

Solution steps

\int_{0}^{1} \frac{1 + x ^{2}}{1 + x ^{4}} d x

Expand

\frac{1 + x ^{2}}{1 + x ^{4}} : \frac{1}{1 + x ^{4}} + \frac{x ^{2}}{1 + x ^{4}}

Apply the Sum Rule:

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

= \int_{0}^{1} \frac{1}{1 + x ^{4}} d x + \int_{0}^{1} \frac{x ^{2}}{1 + x ^{4}} d x

\int_{0}^{1} \frac{1}{1 + x ^{4}} d x = \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{4}} d x = \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

= \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} + \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

Simplify

\frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} + \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} : \frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

= \frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

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What is the integral from 0 to 1 of (1+x^2)/(1+x^4) ?
The integral from 0 to 1 of (1+x^2)/(1+x^4) is (8arctan(sqrt(2)+1)+8arctan(sqrt(2)-1))/(8sqrt(2))