What is the integral of (sqrt(1+(9x)/8)) ?

The integral of (sqrt(1+(9x)/8)) is 1/(27sqrt(2))(8+9x)^{3/2}+C

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integral of (sqrt(1+(9x)/8))

\int (1 + \frac{9 x}{8}) d x

\frac{1}{27 2} (8 + 9 x)^{\frac{3}{2}} + C

Solution steps

\int 1 + \frac{9 x}{8} d x

Apply u-substitution

: \int \frac{8 u}{9} d u

= \int \frac{8 u}{9} d u

Take the constant out:

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

= \frac{8}{9} \cdot \int u d u

Apply the Power Rule

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

= \frac{8}{9} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + \frac{9 x}{8}

= \frac{8}{9} \cdot \frac{2}{3} (1 + \frac{9 x}{8})^{\frac{3}{2}}

Simplify

\frac{8}{9} \cdot \frac{2}{3} (1 + \frac{9 x}{8})^{\frac{3}{2}} : \frac{1}{27 2} (8 + 9 x)^{\frac{3}{2}}

= \frac{1}{27 2} (8 + 9 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27 2} (8 + 9 x)^{\frac{3}{2}} + C

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 2

f (x) = x^{3} + 4 x^{2} + 5 x, at x = - 1

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

(7, - 5), (- 7, - 5)

\frac{d}{d x} (y (e^{x} - 1))

What is the integral of (sqrt(1+(9x)/8)) ?
The integral of (sqrt(1+(9x)/8)) is 1/(27sqrt(2))(8+9x)^{3/2}+C