What is the integral of (2(t-2))/(t(t-4)) ?

The integral of (2(t-2))/(t(t-4)) is ln|t|+ln|t-4|+C

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integral of (2(t-2))/(t(t-4))

\int \frac{2 ( t - 2 )}{t ( t - 4 )} d t

ln ∣ t ∣ + ln ∣ t - 4 ∣ + C

Solution steps

\int \frac{2 ( t - 2 )}{t ( t - 4 )} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{t - 2}{t ( t - 4 )} d t

Expand

\frac{t - 2}{t ( t - 4 )} : \frac{t}{t ( t - 4 )} - \frac{2}{t ( t - 4 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{t}{t ( t - 4 )} d t - \int \frac{2}{t ( t - 4 )} d t)

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

\int \frac{2}{t ( t - 4 )} d t = 2 (- \frac{1}{4} ln ∣ t ∣ + \frac{1}{4} ln ∣ t - 4 ∣)

= 2 (ln ∣ t - 4 ∣ - 2 (- \frac{1}{4} ln ∣ t ∣ + \frac{1}{4} ln ∣ t - 4 ∣))

Simplify

2 (ln ∣ t - 4 ∣ - 2 (- \frac{1}{4} ln ∣ t ∣ + \frac{1}{4} ln ∣ t - 4 ∣)) : ln ∣ t ∣ + ln ∣ t - 4 ∣

= ln ∣ t ∣ + ln ∣ t - 4 ∣

Add a constant to the solution

= ln ∣ t ∣ + ln ∣ t - 4 ∣ + C

What is the integral of (2(t-2))/(t(t-4)) ?
The integral of (2(t-2))/(t(t-4)) is ln|t|+ln|t-4|+C