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

The integral from 0 to 2pi of sqrt(x^4+4x^2) is ((4pi^2+4)^{3/2}-8)/3

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integral from 0 to 2pi of sqrt(x^4+4x^2)

\int_{0}^{2 π} x^{4} + 4 x^{2} d x

\frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

Solution steps

\int_{0}^{2 π} x^{4} + 4 x^{2} d x

Multiply by the conjugate of

x^{4} + 4 x^{2} : \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}}

= \int_{0}^{2 π} \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}} d x

Find undefined (singularity) points:

x = 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

= \int_{0}^{2} \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}} d x + \int_{2}^{2 π} \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}} d x

\int_{0}^{2} \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}} d x = \frac{16 2 - 8}{3}

\int_{2}^{2 π} \frac{x ^{8} - 16 x ^{4}}{x ^{4} - 4 x ^{2}} d x = \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3}

= \frac{16 2 - 8}{3} + \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3}

Simplify

\frac{16 2 - 8}{3} + \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3} : \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

= \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

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What is the integral from 0 to 2pi of sqrt(x^4+4x^2) ?
The integral from 0 to 2pi of sqrt(x^4+4x^2) is ((4pi^2+4)^{3/2}-8)/3