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

The integral of (1-x)/(sqrt(9-4x)) is 1/6 xsqrt(-4x+9)+1/4 sqrt(-4x+9)+C

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

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

\frac{1}{6} x - 4 x + 9 + \frac{1}{4} - 4 x + 9 + C

Solution steps

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

Expand

\frac{1 - x}{9 - 4 x} : \frac{1}{9 - 4 x} - \frac{x}{9 - 4 x}

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

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

\int \frac{x}{9 - 4 x} d x = - \frac{1}{12} (2 x + 9) - 4 x + 9

= - \frac{1}{2} 9 - 4 x - (- \frac{1}{12} (2 x + 9) - 4 x + 9)

Simplify

- \frac{1}{2} 9 - 4 x - (- \frac{1}{12} (2 x + 9) - 4 x + 9) : \frac{1}{6} x - 4 x + 9 + \frac{1}{4} - 4 x + 9

= \frac{1}{6} x - 4 x + 9 + \frac{1}{4} - 4 x + 9

Add a constant to the solution

= \frac{1}{6} x - 4 x + 9 + \frac{1}{4} - 4 x + 9 + C

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What is the integral of (1-x)/(sqrt(9-4x)) ?
The integral of (1-x)/(sqrt(9-4x)) is 1/6 xsqrt(-4x+9)+1/4 sqrt(-4x+9)+C