What is the integral of sqrt((4+9x)/4) ?

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

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integral of sqrt((4+9x)/4)

\int \frac{4 + 9 x}{4} d x

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

Solution steps

\int \frac{4 + 9 x}{4} d x

Simplify

\frac{4 + 9 x}{4} : \frac{1}{2} 4 + 9 x

= \int \frac{1}{2} 4 + 9 x d x

Take the constant out:

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

= \frac{1}{2} \cdot \int 4 + 9 x d x

Apply u-substitution

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

= \frac{1}{2} \cdot \int \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}{2} \cdot \frac{1}{9} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + 9 x

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty (0.9)^{n}

25 y^{^{''}} - 15 y^{^{'}} - 4 y = 0

f (x) = x^{3} (x^{4} - 1)

\frac{d}{d x} (\frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}})

(x^{2} + 4) y = 8

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