What is the integral of sqrt(x^4-4x^2+4) ?

The integral of sqrt(x^4-4x^2+4) is (x^3)/2-(x^3)/6-2x+C

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integral of sqrt(x^4-4x^2+4)

\int x^{4} - 4 x^{2} + 4 d x

\frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x + C

Solution steps

\int x^{4} - 4 x^{2} + 4 d x

Simplify

x^{4} - 4 x^{2} + 4 : (x + 2) (x - 2)

= \int (x + 2) (x - 2) d x

Apply Integration By Parts

= (x + 2) (\frac{x ^{2}}{2} - 2 x) - \int \frac{x ^{2}}{2} - 2 x d x

\int \frac{x ^{2}}{2} - 2 x d x = \frac{x ^{3}}{6} - \frac{x ^{2}}{2}

= (x + 2) (\frac{x ^{2}}{2} - 2 x) - (\frac{x ^{3}}{6} - \frac{x ^{2}}{2})

Simplify

(x + 2) (\frac{x ^{2}}{2} - 2 x) - (\frac{x ^{3}}{6} - \frac{x ^{2}}{2}) : \frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x

= \frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x

Add a constant to the solution

= \frac{x ^{3}}{2} - \frac{x ^{3}}{6} - 2 x + C

What is the integral of sqrt(x^4-4x^2+4) ?
The integral of sqrt(x^4-4x^2+4) is (x^3)/2-(x^3)/6-2x+C