What is the integral of (1-3/x+y) ?

The integral of (1-3/x+y) is x-3ln|x|+yx+C

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integral of (1-3/x+y)

\int (1 - \frac{3}{x} + y) d x

x - 3 ln ∣ x ∣ + y x + C

Solution steps

\int 1 - \frac{3}{x} + y 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}{x} d x + \int y d x

\int 1 d x = x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int y d x = y x

= x - 3 ln ∣ x ∣ + y x

Add a constant to the solution

= x - 3 ln ∣ x ∣ + y x + C

\frac{d y}{d t} = 0.07 y + 200

x \to \infty lim (- 8 x^{6})

\frac{d}{d x} (e^{x^{2} - 2})

\int - 8 x + 10 y - 3 d y

y = lo g_{3} (x)