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

The integral of sqrt(x^2-2x^4) is -1/6 (1-2x^2)^{3/2}+C

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

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

- \frac{1}{6} (1 - 2 x^{2})^{\frac{3}{2}} + C

Solution steps

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

Simplify

x^{2} - 2 x^{4} : x^{2} 1 - 2 x^{2}

= \int x^{2} 1 - 2 x^{2} d x

x^{2} = (x),

assuming

x \geq 0

= \int x 1 - 2 x^{2} d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 - 2 x^{2}

= - \frac{1}{4} \cdot \frac{2}{3} (1 - 2 x^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (1 - 2 x^{2})^{\frac{3}{2}} : - \frac{1}{6} (1 - 2 x^{2})^{\frac{3}{2}}

= - \frac{1}{6} (1 - 2 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (1 - 2 x^{2})^{\frac{3}{2}} + C

n = 0 \sum \infty \frac{2}{e ^{n}}

\frac{\partial}{\partial x} (sin (3 x - 4 y))

\frac{d}{d x} (cos (9))

\int_{0}^{1} \frac{1}{x ( 2 x + 5 )} d x

\frac{\partial}{\partial x} (tan^{2} (x + y))

What is the integral of sqrt(x^2-2x^4) ?
The integral of sqrt(x^2-2x^4) is -1/6 (1-2x^2)^{3/2}+C