What is the integral of (1-3x)/(2sqrt(1-x)) ?

The integral of (1-3x)/(2sqrt(1-x)) is (x+1)sqrt(-x+1)+C

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integral of (1-3x)/(2sqrt(1-x))

\int \frac{1 - 3 x}{2 1 - x} d x

(x + 1) - x + 1 + C

Solution steps

\int \frac{1 - 3 x}{2 1 - x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1 - 3 x}{1 - x} d x

Expand

\frac{1 - 3 x}{1 - x} : \frac{1}{1 - x} - \frac{3 x}{1 - x}

= \frac{1}{2} \cdot \int \frac{1}{1 - x} - \frac{3 x}{1 - 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

= \frac{1}{2} (\int \frac{1}{1 - x} d x - \int \frac{3 x}{1 - x} d x)

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

\int \frac{3 x}{1 - x} d x = - 6 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

= \frac{1}{2} (- 2 1 - x - (- 6 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x)))

Simplify

\frac{1}{2} (- 2 1 - x - (- 6 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x))) : (x + 1) - x + 1

= (x + 1) - x + 1

Add a constant to the solution

= (x + 1) - x + 1 + C

\int 7 x e^{3 x} d x

\int 484 - x^{2} d x

\frac{\partial}{\partial x} (x e^{x^{2} + 2 y^{2}})

f (x) = 3 x^{3} - x^{2} + 9, (2, 29)

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

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