What is the integral from 0 to 1 of 2x^3(3x^4+2)^6 ?

The integral from 0 to 1 of 2x^3(3x^4+2)^6 is 25999/14

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integral from 0 to 1 of 2x^3(3x^4+2)^6

\int_{0}^{1} 2 x^{3} (3 x^{4} + 2)^{6} d x

\frac{25999}{14}

Decimal

1857.07142 \dots

Solution steps

\int_{0}^{1} 2 x^{3} (3 x^{4} + 2)^{6} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} x^{3} (3 x^{4} + 2)^{6} d x

Multiply by the conjugate of

3 x^{4} + 2 : \frac{9 x ^{8} - 4}{3 x ^{4} - 2}

= 2 \cdot \int_{0}^{1} x^{3} (\frac{9 x ^{8} - 4}{3 x ^{4} - 2})^{6} d x

Find undefined (singularity) points:

x = 4 \frac{2}{3}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= 2 (\int_{0}^{4 \frac{2}{3}} x^{3} (\frac{9 x ^{8} - 4}{3 x ^{4} - 2})^{6} d x + \int_{4 \frac{2}{3}}^{1} x^{3} (\frac{9 x ^{8} - 4}{3 x ^{4} - 2})^{6} d x)

\int_{0}^{4 \frac{2}{3}} x^{3} (\frac{9 x ^{8} - 4}{3 x ^{4} - 2})^{6} d x = \frac{4064}{21}

\int_{4 \frac{2}{3}}^{1} x^{3} (\frac{9 x ^{8} - 4}{3 x ^{4} - 2})^{6} d x = \frac{61741}{84}

= 2 (\frac{4064}{21} + \frac{61741}{84})

Simplify

2 (\frac{4064}{21} + \frac{61741}{84}) : \frac{25999}{14}

= \frac{25999}{14}

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What is the integral from 0 to 1 of 2x^3(3x^4+2)^6 ?
The integral from 0 to 1 of 2x^3(3x^4+2)^6 is 25999/14