What is the integral from 1 to 2x of 1/(1+t^3) ?

The integral from 1 to 2x of 1/(1+t^3) is

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integral from 1 to 2x of 1/(1+t^3)

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

\frac{1}{3} (ln ∣ 2 x + 1 ∣ - ln (2)) + \frac{1}{3} (- \frac{1}{2} ln (\frac{16 x ^{2} - 8 x + 4}{4}) + 3 (arctan (\frac{4 x - 1}{3}) - \frac{π}{6}))

Solution steps

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

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

\int_{1}^{2 x} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{3} (- \frac{1}{2} ln (\frac{16 x ^{2} - 8 x + 4}{4}) + 3 (arctan (\frac{4 x - 1}{3}) - \frac{π}{6}))

= \frac{1}{3} (ln ∣ 2 x + 1 ∣ - ln (2)) + \frac{1}{3} (- \frac{1}{2} ln (\frac{16 x ^{2} - 8 x + 4}{4}) + 3 (arctan (\frac{4 x - 1}{3}) - \frac{π}{6}))

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What is the integral from 1 to 2x of 1/(1+t^3) ?
The integral from 1 to 2x of 1/(1+t^3) is