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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\int \frac{x ^{3}}{x + 3} d x

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

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

\int \frac{3}{x ^{3} + 1} d x

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