What is the integral of sqrt(9x^4+4x^2) ?

The integral of sqrt(9x^4+4x^2) is 1/27 (9x^2+4)^{3/2}+C

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integral of sqrt(9x^4+4x^2)

\int 9 x^{4} + 4 x^{2} d x

\frac{1}{27} (9 x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int 9 x^{4} + 4 x^{2} d x

Factor

9 x^{4} + 4 x^{2} : x^{2} 9 x^{2} + 4

= \int x^{2} 9 x^{2} + 4 d x

x^{2} = (x),

assuming

x \geq 0

= \int x 9 x^{2} + 4 d x

Apply u-substitution

: \int \frac{u}{18} d u

= \int \frac{u}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{18} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 x^{2} + 4

= \frac{1}{18} \cdot \frac{2}{3} (9 x^{2} + 4)^{\frac{3}{2}}

Simplify

\frac{1}{18} \cdot \frac{2}{3} (9 x^{2} + 4)^{\frac{3}{2}} : \frac{1}{27} (9 x^{2} + 4)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (9 x^{2} + 4)^{\frac{3}{2}} + C

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

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (9) = 9

\frac{d}{d x} (\frac{x ^{2}}{( 3 + x )})

\frac{\partial}{\partial x} (x^{7})

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 9 y^{3}))

What is the integral of sqrt(9x^4+4x^2) ?
The integral of sqrt(9x^4+4x^2) is 1/27 (9x^2+4)^{3/2}+C