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

The integral from 0 to x of sqrt(x^2+x^4) is ((1+x^2)^{3/2}-1)/3

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

\int_{0}^{x} x^{2} + x^{4} d x

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

Solution steps

\int_{0}^{x} x^{2} + x^{4} d x

Simplify

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

= \int_{0}^{x} x^{2} 1 + x^{2} d x

x^{2} = (x),

assuming

x \geq 0

= \int_{0}^{x} x 1 + x^{2} d x

Apply u-substitution

: \int_{1}^{1 + x^{2}} u^{2} d u

= \int_{1}^{1 + x^{2}} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{1 + x^{2}}

= [\frac{u ^{3}}{3}]_{1}^{1 + x^{2}}

Compute the boundaries:

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

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

Simplify

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

\frac{d}{d x} (a x^{2} e^{3 x})

\int 2 cos (x) cos (2 x) d x

\frac{d}{d x} (arctan (0))

\frac{\partial}{\partial y} (\frac{x}{x + y})

\frac{\partial}{\partial y} (- \frac{1}{4} (y^{2} + x^{2}))

What is the integral from 0 to x of sqrt(x^2+x^4) ?
The integral from 0 to x of sqrt(x^2+x^4) is ((1+x^2)^{3/2}-1)/3