What is the integral of x/(sqrt(4-3x^2)) ?

The integral of x/(sqrt(4-3x^2)) is -1/3 sqrt(4-3x^2)+C

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integral of x/(sqrt(4-3x^2))

\int \frac{x}{4 - 3 x ^{2}} d x

- \frac{1}{3} 4 - 3 x^{2} + C

Solution steps

\int \frac{x}{4 - 3 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{6 u} d u

= \int - \frac{1}{6 u} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 - 3 x^{2}

= - \frac{1}{6} \cdot 2 (4 - 3 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{6} \cdot 2 (4 - 3 x^{2})^{\frac{1}{2}} : - \frac{1}{3} 4 - 3 x^{2}

= - \frac{1}{3} 4 - 3 x^{2}

Add a constant to the solution

= - \frac{1}{3} 4 - 3 x^{2} + C

x \to 0 lim (\frac{x + 97}{x})

x y^{^{'}} - 2 y = 2 x^{2} + 2 x + 2

g (x) = 3 e^{- 2 x}, (0, 3)

f (x) = \frac{( x ^{2} - 1 )}{x ^{2} + x + 1}

\frac{d}{d x} (\frac{1}{3} ln (x))

What is the integral of x/(sqrt(4-3x^2)) ?
The integral of x/(sqrt(4-3x^2)) is -1/3 sqrt(4-3x^2)+C