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

The integral of (x+1)/(x-3) is 4ln|x-3|+x+C

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

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

4 ln ∣ x - 3 ∣ + x + C

Solution steps

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

\frac{x + 1}{x - 3} = \frac{4}{x - 3} + 1

= \int \frac{4}{x - 3} + 1 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{4}{x - 3} d x + \int 1 d x

\int \frac{4}{x - 3} d x = 4 ln ∣ x - 3 ∣

\int 1 d x = x

= 4 ln ∣ x - 3 ∣ + x

Add a constant to the solution

= 4 ln ∣ x - 3 ∣ + x + C

\frac{d y}{d x} = e^{(x - y)}

f (x) = cos^{2} (2 x)

x \to - 1 - lim (4 - x)

x \to π lim (7 cot (x))

\frac{d}{d t} (e^{- t} (3 cos (\frac{t}{2}) + 2 sin (\frac{t}{2})))

What is the integral of (x+1)/(x-3) ?
The integral of (x+1)/(x-3) is 4ln|x-3|+x+C