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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

\int \frac{3 t ^{3}}{1 + t ^{3}} 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 (1 + t^{3}) - (- 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}^{x}

Simplify

[t ln (1 + t^{3}) - (- 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}^{x} : [t ln (1 + t^{3}) + 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}^{x}

= [t ln (1 + t^{3}) + 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}^{x}

Compute the boundaries:

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

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

Simplify

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

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