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

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

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

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

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

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

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 t^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x + y - 320)

x \to 4 lim (x - 2)

\frac{d}{d x} (e^{x^{2}} y) = x

\int_{0}^{7} x + 1 d x

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

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