What is the integral from 16 to x of sqrt(1+9t) ?

The integral from 16 to x of sqrt(1+9t) is (2(1+9x)^{3/2}-290sqrt(145))/(27)

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integral from 16 to x of sqrt(1+9t)

\int_{16}^{x} 1 + 9 t d t

\frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 290 145}{27}

Solution steps

\int_{16}^{x} 1 + 9 t d t

Apply u-substitution

: \int_{145}^{1 + 9 x} \frac{u}{9} d u

= \int_{145}^{1 + 9 x} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{145}^{1 + 9 x} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{145}^{1 + 9 x}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{145}^{1 + 9 x}

Compute the boundaries:

\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{290 145}{3}

= \frac{1}{9} (\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{290 145}{3})

Simplify

= \frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 290 145}{27}

\frac{d y}{d x} = - 5 sin (4 x)

\int \frac{9}{x ^{2} + 3 x - 18} d x

\int_{- 3}^{1} ∣ 3 x + 3 ∣ d x

x^{2} y^{^{'}} = 2 y^{2} - x^{2}

What is the integral from 16 to x of sqrt(1+9t) ?
The integral from 16 to x of sqrt(1+9t) is (2(1+9x)^{3/2}-290sqrt(145))/(27)