What is the integral from 1 to x^3 of 1/(1+t^4) ?

The integral from 1 to x^3 of 1/(1+t^4) is

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integral from 1 to x^3 of 1/(1+t^4)

\int_{1}^{x^{3}} \frac{1}{1 + t ^{4}} d t

\frac{1}{2} (\frac{1}{2 2} (ln 2 x^{6} + 22 x^{3} + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (2 x^{3} + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{6} - 22 x^{3} + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (2 x^{3} - 1) - arctan (2 - 1)))

Solution steps

\int_{1}^{x^{3}} \frac{1}{1 + t ^{4}} d t

Take the partial fraction of

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

= \int_{1}^{x^{3}} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} + \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t

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}^{x^{3}} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t + \int_{1}^{x^{3}} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t

\int_{1}^{x^{3}} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{6} + 22 x^{3} + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (2 x^{3} + 1) - arctan (2 + 1)))

\int_{1}^{x^{3}} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{6} - 22 x^{3} + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (2 x^{3} - 1) - arctan (2 - 1)))

= \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{6} + 22 x^{3} + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (2 x^{3} + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{6} - 22 x^{3} + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (2 x^{3} - 1) - arctan (2 - 1)))

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What is the integral from 1 to x^3 of 1/(1+t^4) ?
The integral from 1 to x^3 of 1/(1+t^4) is