What is the integral from 0 to 2x of ln(t^3+1) ?

The integral from 0 to 2x of ln(t^3+1) is

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integral from 0 to 2x of ln(t^3+1)

\int_{0}^{2 x} ln (t^{3} + 1) d t

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

Solution steps

\int_{0}^{2 x} ln (t^{3} + 1) d t

Apply Integration By Parts

: [t ln (t^{3} + 1) - \int \frac{3 t ^{3}}{t ^{3} + 1} d t]_{0}^{2 x}

= [t ln (t^{3} + 1) - \int \frac{3 t ^{3}}{t ^{3} + 1} d t]_{0}^{2 x}

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

= [t ln (t^{3} + 1) - (- ln ∣ t + 1 ∣ - \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) + 3 t)]_{0}^{2 x}

Simplify

[t ln (t^{3} + 1) - (- ln ∣ t + 1 ∣ - \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) + 3 t)]_{0}^{2 x} : [t ln (t^{3} + 1) + ln ∣ t + 1 ∣ + \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) - 3 t]_{0}^{2 x}

= [t ln (t^{3} + 1) + ln ∣ t + 1 ∣ + \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) - 3 t]_{0}^{2 x}

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to 2x of ln(t^3+1) ?
The integral from 0 to 2x of ln(t^3+1) is