What is the integral from 0 to 1 of 40x(2x^2+1)^4 ?

The integral from 0 to 1 of 40x(2x^2+1)^4 is 484

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integral from 0 to 1 of 40x(2x^2+1)^4

\int_{0}^{1} 40 x (2 x^{2} + 1)^{4} d x

484

Solution steps

\int_{0}^{1} 40 x (2 x^{2} + 1)^{4} d x

Take the constant out:

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

= 40 \cdot \int_{0}^{1} x (2 x^{2} + 1)^{4} d x

Multiply by the conjugate of

2 x^{2} + 1 : \frac{4 x ^{4} - 1}{2 x ^{2} - 1}

= 40 \cdot \int_{0}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x

Find undefined (singularity) points:

x = \frac{1}{2}

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

= 40 (\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x + \int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x)

\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{31}{20}

\int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{211}{20}

= 40 (\frac{31}{20} + \frac{211}{20})

Simplify

40 (\frac{31}{20} + \frac{211}{20}) : 484

= 484

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What is the integral from 0 to 1 of 40x(2x^2+1)^4 ?
The integral from 0 to 1 of 40x(2x^2+1)^4 is 484