What is the integral of 1-3/(sqrt(14-x)) ?

The integral of 1-3/(sqrt(14-x)) is x+6sqrt(14-x)+C

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integral of 1-3/(sqrt(14-x))

\int 1 - \frac{3}{14 - x} d x

x + 6 14 - x + C

Solution steps

\int 1 - \frac{3}{14 - 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 1 d x - \int \frac{3}{14 - x} d x

\int 1 d x = x

\int \frac{3}{14 - x} d x = - 6 14 - x

= x - (- 6 14 - x)

Simplify

= x + 6 14 - x

Add a constant to the solution

= x + 6 14 - x + C

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

\frac{d y}{d t} + y = 7 t

(7, - 2), (3, 5)

x \to 0 lim (\frac{7 x ^{2} - 6 x}{x})

\frac{d}{d x} (8 x^{3} - cos (4 x))

What is the integral of 1-3/(sqrt(14-x)) ?
The integral of 1-3/(sqrt(14-x)) is x+6sqrt(14-x)+C