What is the integral of (1-2x)/(sqrt(16-9x^2)) ?

The integral of (1-2x)/(sqrt(16-9x^2)) is 1/3 arcsin(3/4 x)+2/9 sqrt(16-9x^2)+C

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integral of (1-2x)/(sqrt(16-9x^2))

\int \frac{1 - 2 x}{16 - 9 x ^{2}} d x

\frac{1}{3} arcsin (\frac{3}{4} x) + \frac{2}{9} 16 - 9 x^{2} + C

Solution steps

\int \frac{1 - 2 x}{16 - 9 x ^{2}} d x

Expand

\frac{1 - 2 x}{16 - 9 x ^{2}} : \frac{1}{16 - 9 x ^{2}} - \frac{2 x}{16 - 9 x ^{2}}

= \int \frac{1}{16 - 9 x ^{2}} - \frac{2 x}{16 - 9 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}{16 - 9 x ^{2}} d x - \int \frac{2 x}{16 - 9 x ^{2}} d x

\int \frac{1}{16 - 9 x ^{2}} d x = \frac{1}{3} arcsin (\frac{3}{4} x)

\int \frac{2 x}{16 - 9 x ^{2}} d x = - \frac{2}{9} 16 - 9 x^{2}

= \frac{1}{3} arcsin (\frac{3}{4} x) - (- \frac{2}{9} 16 - 9 x^{2})

Simplify

= \frac{1}{3} arcsin (\frac{3}{4} x) + \frac{2}{9} 16 - 9 x^{2}

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{4} x) + \frac{2}{9} 16 - 9 x^{2} + C

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What is the integral of (1-2x)/(sqrt(16-9x^2)) ?
The integral of (1-2x)/(sqrt(16-9x^2)) is 1/3 arcsin(3/4 x)+2/9 sqrt(16-9x^2)+C