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

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

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

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

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

Solution steps

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

Simplify

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

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

t^{2} = (t),

assuming

t \geq 0

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

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

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

Simplify

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

\frac{d y}{d x} = y^{2} - 4

\frac{\partial}{\partial x} (sin (5 π x + 2 y))

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

f (x) = x^{4} - 20 x^{2} + 64, at x = 2

x \to \frac{5 π}{2} lim (sec (x))

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