What is the integral of (18x)/(sqrt(1-9x^2)) ?

The integral of (18x)/(sqrt(1-9x^2)) is -2sqrt(1-9x^2)+C

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integral of (18x)/(sqrt(1-9x^2))

\int \frac{18 x}{1 - 9 x ^{2}} d x

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

Solution steps

\int \frac{18 x}{1 - 9 x ^{2}} d x

Take the constant out:

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

= 18 \cdot \int \frac{x}{1 - 9 x ^{2}} d x

Apply u-substitution

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

= 18 \cdot \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

= 18 (- \frac{1}{18} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

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

Substitute back

u = 1 - 9 x^{2}

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

Simplify

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

= - 2 1 - 9 x^{2}

Add a constant to the solution

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

\int 3 x (6 - x^{2}) d x

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

\int y^{3} (2 y^{2} - 3) d y

\frac{\partial}{\partial x} (2 y + y e^{x})

\frac{d x}{d t} = - 3 x

What is the integral of (18x)/(sqrt(1-9x^2)) ?
The integral of (18x)/(sqrt(1-9x^2)) is -2sqrt(1-9x^2)+C