What is the integral of x/(sqrt(9x^2-4)) ?

The integral of x/(sqrt(9x^2-4)) is 1/9 sqrt(9x^2-4)+C

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integral of x/(sqrt(9x^2-4))

\int \frac{x}{9 x ^{2} - 4} d x

\frac{1}{9} 9 x^{2} - 4 + C

Solution steps

\int \frac{x}{9 x ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{18 u} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 9 x^{2} - 4

= \frac{1}{18} \cdot 2 (9 x^{2} - 4)^{\frac{1}{2}}

Simplify

\frac{1}{18} \cdot 2 (9 x^{2} - 4)^{\frac{1}{2}} : \frac{1}{9} 9 x^{2} - 4

= \frac{1}{9} 9 x^{2} - 4

Add a constant to the solution

= \frac{1}{9} 9 x^{2} - 4 + C

y^{^{''}} + 6 y^{^{'}} + 34 y = 68, y (0) = 5, y^{^{'}} (0) = 7

\int x^{0.813} d x

x y + y^{^{'}} = 18 x

\frac{d}{d x} (\frac{( g ( x ) ) ^{2}}{h ( x )})

\frac{d}{d x} (ln (\frac{2 - x}{2 + x}))

What is the integral of x/(sqrt(9x^2-4)) ?
The integral of x/(sqrt(9x^2-4)) is 1/9 sqrt(9x^2-4)+C