What is the integral from 0 to 0.3 of xln(1+x^3) ?

The integral from 0 to 0.3 of xln(1+x^3) is 0.00048…

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integral from 0 to 0.3 of xln(1+x^3)

\int_{0}^{0.3} x ln (1 + x^{3}) d x

0.00048 \dots

Solution steps

\int_{0}^{0.3} x ln (1 + x^{3}) d x

Apply Integration By Parts

: [\frac{1}{2} (x^{2} ln (1 + x^{3}) - 2 \cdot \int \frac{3 x ^{4}}{2 ( 1 + x ^{3} )} d x)]_{0}^{0.3}

= [\frac{1}{2} (x^{2} ln (1 + x^{3}) - 2 \cdot \int \frac{3 x ^{4}}{2 ( 1 + x ^{3} )} d x)]_{0}^{0.3}

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

= [\frac{1}{2} (x^{2} ln (1 + x^{3}) - 2 (\frac{3 x ^{2}}{4} + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{4} ln 4 x^{2} - 4 x + 4 - \frac{3}{2} arctan (\frac{2 x - 1}{3})))]_{0}^{0.3}

Simplify

[\frac{1}{2} (x^{2} ln (1 + x^{3}) - 2 (\frac{3 x ^{2}}{4} + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{4} ln 4 x^{2} - 4 x + 4 - \frac{3}{2} arctan (\frac{2 x - 1}{3})))]_{0}^{0.3} : [\frac{1}{4} (2 x^{2} ln (1 + x^{3}) - 3 x^{2} - 2 ln ∣ x + 1 ∣ + ln 4 x^{2} - 4 x + 4 + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}))]_{0}^{0.3}

= [\frac{1}{4} (2 x^{2} ln (1 + x^{3}) - 3 x^{2} - 2 ln ∣ x + 1 ∣ + ln 4 x^{2} - 4 x + 4 + 3.46410 \dots arctan (\frac{2 x - 1}{1.73205 \dots}))]_{0}^{0.3}

Compute the boundaries:

0.00048 \dots

= 0.00048 \dots

\int_{1}^{2} (3 x^{3} + \frac{6}{x ^{2}}) d x

\int_{- 2}^{2} x^{5} d x

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

\int_{0}^{5} 6

\int_{1}^{2} (4 x^{3} - 6) d x

What is the integral from 0 to 0.3 of xln(1+x^3) ?
The integral from 0 to 0.3 of xln(1+x^3) is 0.00048…