What is the integral from 0 to 0.1 of ln(1+x^3) ?

The integral from 0 to 0.1 of ln(1+x^3) is 0.00002…

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integral from 0 to 0.1 of ln(1+x^3)

\int_{0}^{0.1} ln (1 + x^{3}) d x

0.00002 \dots

Solution steps

\int_{0}^{0.1} ln (1 + x^{3}) d x

Apply Integration By Parts

: [x ln (1 + x^{3}) - \int \frac{3 x ^{3}}{1 + x ^{3}} d x]_{0}^{0.1}

= [x ln (1 + x^{3}) - \int \frac{3 x ^{3}}{1 + x ^{3}} d x]_{0}^{0.1}

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

= [x ln (1 + x^{3}) - (- ln ∣ x + 1 ∣ - \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + 3 x)]_{0}^{0.1}

Simplify

[x ln (1 + x^{3}) - (- ln ∣ x + 1 ∣ - \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + 3 x)]_{0}^{0.1} : [\frac{1}{2} (2 x ln (1 + x^{3}) + 2 ln ∣ x + 1 ∣ - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}) - 6 x)]_{0}^{0.1}

= [\frac{1}{2} (2 x ln (1 + x^{3}) + 2 ln ∣ x + 1 ∣ - ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}) - 6 x)]_{0}^{0.1}

Compute the boundaries:

0.00002 \dots

= 0.00002 \dots

\int_{0}^{1} x^{2} - x^{6} d x

\int_{x^{2}}^{x} 1 d y

\int_{- 1}^{1} x^{4} 1 - x^{2} d x

\int_{- 4}^{5} - x^{2} + x + 20 d x

\int_{4}^{6} x^{2} + 2 x - 2 d x

What is the integral from 0 to 0.1 of ln(1+x^3) ?
The integral from 0 to 0.1 of ln(1+x^3) is 0.00002…