What is the integral of (8t-5)/(t+1) ?

The integral of (8t-5)/(t+1) is 8(t+1)-13ln|t+1|+C

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integral of (8t-5)/(t+1)

\int \frac{8 t - 5}{t + 1} d t

8 (t + 1) - 13 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{8 t - 5}{t + 1} d t

Apply u-substitution

= \int \frac{8 u - 13}{u} d u

Expand

\frac{8 u - 13}{u} : 8 - \frac{13}{u}

= \int 8 - \frac{13}{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 8 d u - \int \frac{13}{u} d u

\int 8 d u = 8 u

\int \frac{13}{u} d u = 13 ln ∣ u ∣

= 8 u - 13 ln ∣ u ∣

Substitute back

u = t + 1

= 8 (t + 1) - 13 ln ∣ t + 1 ∣

Add a constant to the solution

= 8 (t + 1) - 13 ln ∣ t + 1 ∣ + C

What is the integral of (8t-5)/(t+1) ?
The integral of (8t-5)/(t+1) is 8(t+1)-13ln|t+1|+C