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

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

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integral of (1/x+1/(3-x))

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

ln ∣ x ∣ - ln ∣ 3 - x ∣ + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{3 - x} d x = - ln ∣ 3 - x ∣

= ln ∣ x ∣ - ln ∣ 3 - x ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ 3 - x ∣ + C

x \to 0 lim (sin (5 x) csc (3 x))

0.02 (1 + 3 cos (t))

\frac{d y}{d t} = 25 - 10 t - 2 y

2 (x + 1), 3 (x + 1), x = 3

\int_{- \frac{π}{6}}^{0} 6 sec^{3} (x) d x

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