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

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

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

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

4 ln ∣ t - 3 ∣ + t + C

Solution steps

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

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

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

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}{t - 3} d t + \int 1 d t

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

\int 1 d t = t

= 4 ln ∣ t - 3 ∣ + t

Add a constant to the solution

= 4 ln ∣ t - 3 ∣ + t + C

cot (5 x)

\frac{d}{d x} (- 2 sin (2 x) + \frac{1}{x})

y = x - x^{3}

\int e^{- 2 \cdot π \cdot x} \cdot cos (π \cdot x) d x

n = 1 \sum \infty \frac{1}{4 n ^{4}}

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