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

The integral of ((1-x))/(sqrt(9-4x^2)) is 1/2 arcsin(2/3 x)+1/4 sqrt(9-4x^2)+C

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integral of ((1-x))/(sqrt(9-4x^2))

\int \frac{( 1 - x )}{9 - 4 x ^{2}} d x

\frac{1}{2} arcsin (\frac{2}{3} x) + \frac{1}{4} 9 - 4 x^{2} + C

Solution steps

\int \frac{1 - x}{9 - 4 x ^{2}} d x

Simplify

x^{2} : xx

= \int \frac{1 - x}{9 - 4 xx} d x

Expand

\frac{1 - x}{9 - 4 xx} : \frac{1}{9 - 4 x ^{2}} - \frac{x}{9 - 4 x ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{9 - 4 x ^{2}} d x = \frac{1}{2} arcsin (\frac{2}{3} x)

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

= \frac{1}{2} arcsin (\frac{2}{3} x) - (- \frac{1}{4} 9 - 4 x^{2})

Simplify

= \frac{1}{2} arcsin (\frac{2}{3} x) + \frac{1}{4} 9 - 4 x^{2}

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{2}{3} x) + \frac{1}{4} 9 - 4 x^{2} + C

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What is the integral of ((1-x))/(sqrt(9-4x^2)) ?
The integral of ((1-x))/(sqrt(9-4x^2)) is 1/2 arcsin(2/3 x)+1/4 sqrt(9-4x^2)+C