What is the integral of (5t-3)/(t+1) ?

The integral of (5t-3)/(t+1) is 5(t+1)-8ln|t+1|+C

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integral of (5t-3)/(t+1)

\int \frac{5 t - 3}{t + 1} d t

5 (t + 1) - 8 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{5 t - 3}{t + 1} d t

Apply u-substitution

= \int \frac{5 u - 8}{u} d u

Expand

\frac{5 u - 8}{u} : 5 - \frac{8}{u}

= \int 5 - \frac{8}{u} d u

Apply the Sum Rule:

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

= \int 5 d u - \int \frac{8}{u} d u

\int 5 d u = 5 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 5 u - 8 ln ∣ u ∣

Substitute back

u = t + 1

= 5 (t + 1) - 8 ln ∣ t + 1 ∣

Add a constant to the solution

= 5 (t + 1) - 8 ln ∣ t + 1 ∣ + C

What is the integral of (5t-3)/(t+1) ?
The integral of (5t-3)/(t+1) is 5(t+1)-8ln|t+1|+C