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

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

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integral of x/(sqrt(1-9x^2))

\int \frac{x}{1 - 9 x ^{2}} d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

x \to 4 lim (\frac{∣ 4 - x ∣}{4 - x})

\int (2 x^{- 1}) d x

2 y^{^{''}} + y^{^{'}} - 3 y = 0

\int (2 x - 5)^{3} d x

y^{^{'}} = \frac{y}{4 x}

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