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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{1 + t ^{3}} : \frac{1}{3 ( t + 1 )} + \frac{- t + 2}{3 ( t ^{2} - t + 1 )}

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

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

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

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

\int_{- x}^{x} x + y d y

\int_{0}^{1} - x^{2} - x^{3} + 2 x d x

\int_{- 2}^{x} (28 t^{2} - 27 t) d t

\int_{1}^{5} (\frac{8}{x} - e^{- x}) d x

\int_{0}^{5} (2 t - 1)^{2} d t

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