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

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

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

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

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

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

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

Substitute back

u = 4 + 9 t^{2}

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

Simplify

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

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

Add a constant to the solution

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

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

d er i v a t i v e \frac{x ^{4}}{4}

t an g e n t f (x) = 9 e^{x} cos (x), at x = 0

\frac{\partial}{\partial x} (\frac{1}{1 + e ^{- x}})

\int \frac{- x}{1 + x} d x

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