What is the integral from 0 to 0.1 of 1/(1+x^4) ?

The integral from 0 to 0.1 of 1/(1+x^4) is 0.09999…

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integral from 0 to 0.1 of 1/(1+x^4)

\int_{0}^{0.1} \frac{1}{1 + x ^{4}} d x

0.09999 \dots

Solution steps

\int_{0}^{0.1} \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_{0}^{0.1} \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_{0}^{0.1} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x + \int_{0}^{0.1} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

\int_{0}^{0.1} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = 0.04824 \dots

\int_{0}^{0.1} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = 0.05175 \dots

= 0.04824 \dots + 0.05175 \dots

Simplify

= 0.09999 \dots

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What is the integral from 0 to 0.1 of 1/(1+x^4) ?
The integral from 0 to 0.1 of 1/(1+x^4) is 0.09999…