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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{8 u} d u

= \int - \frac{1}{8 u} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 9 - 4 x^{2}

= - \frac{1}{8} \cdot 2 (9 - 4 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{8} \cdot 2 (9 - 4 x^{2})^{\frac{1}{2}} : - \frac{1}{4} 9 - 4 x^{2}

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

Add a constant to the solution

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

\int \frac{( 1 + 6 ln ( x ) ) ^{2}}{x} d x

\frac{d}{d x} (\frac{b}{x})

\frac{d y}{d x} = \frac{x ^{2} y ^{2}}{1 + x}

\int_{0}^{5} 2 x^{2} e^{- x} d x

3500 (1 - \frac{1}{50} t)^{2}

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