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

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

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

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

\frac{1}{3} (ln x^{2} + 1 - ln ∣ 3 x + 1 ∣) + \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{4} - 4 x ^{2} + 4}{∣ 36 x ^{2} - 12 x + 4 ∣}) + 3 (arctan (\frac{2 x ^{2} - 1}{3}) - arctan (23 x - \frac{1}{3})))

Solution steps

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

Take the partial fraction of

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

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

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

\int_{3 x}^{x^{2}} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{4} - 4 x ^{2} + 4}{∣ 36 x ^{2} - 12 x + 4 ∣}) + 3 (arctan (\frac{2 x ^{2} - 1}{3}) - arctan (23 x - \frac{1}{3})))

= \frac{1}{3} (ln x^{2} + 1 - ln ∣ 3 x + 1 ∣) + \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{4} - 4 x ^{2} + 4}{∣ 36 x ^{2} - 12 x + 4 ∣}) + 3 (arctan (\frac{2 x ^{2} - 1}{3}) - arctan (23 x - \frac{1}{3})))

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