What is the integral of sqrt(100t^2+36t^4) ?

The integral of sqrt(100t^2+36t^4) is 2/27 (9t^2+25)^{3/2}+C

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integral of sqrt(100t^2+36t^4)

\int 100 t^{2} + 36 t^{4} d t

\frac{2}{27} (9 t^{2} + 25)^{\frac{3}{2}} + C

Solution steps

\int 100 t^{2} + 36 t^{4} d t

Factor

100 t^{2} + 36 t^{4} : 2 t^{2} 25 + 9 t^{2}

= \int 2 t^{2} 25 + 9 t^{2} d t

Take the constant out:

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

= 2 \cdot \int t^{2} 25 + 9 t^{2} d t

t^{2} = (t),

assuming

t \geq 0

= 2 \cdot \int t 25 + 9 t^{2} d t

Apply u-substitution

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

= 2 \cdot \int \frac{u}{18} d u

Take the constant out:

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

= 2 \cdot \frac{1}{18} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 25 + 9 t^{2}

= 2 \cdot \frac{1}{18} \cdot \frac{2}{3} (25 + 9 t^{2})^{\frac{3}{2}}

Simplify

2 \cdot \frac{1}{18} \cdot \frac{2}{3} (25 + 9 t^{2})^{\frac{3}{2}} : \frac{2}{27} (9 t^{2} + 25)^{\frac{3}{2}}

= \frac{2}{27} (9 t^{2} + 25)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{27} (9 t^{2} + 25)^{\frac{3}{2}} + C

9 \frac{d}{d x} (\frac{d y}{d x}) + 10 y = 0

x \to - 3 - lim (\frac{x}{x + 3})

x \to 0 lim ((x) (ln (x))^{2})

\frac{d}{d x} (sin (x) x)

sin (a x)

What is the integral of sqrt(100t^2+36t^4) ?
The integral of sqrt(100t^2+36t^4) is 2/27 (9t^2+25)^{3/2}+C