What is the integral of (9t-4)/(t+1) ?

The integral of (9t-4)/(t+1) is 9(t+1)-13ln|t+1|+C

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integral of (9t-4)/(t+1)

\int \frac{9 t - 4}{t + 1} d t

9 (t + 1) - 13 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{9 t - 4}{t + 1} d t

Apply u-substitution

: \int \frac{9 u - 13}{u} d u

= \int \frac{9 u - 13}{u} d u

Expand

\frac{9 u - 13}{u} : 9 - \frac{13}{u}

= \int 9 - \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 9 d u - \int \frac{13}{u} d u

\int 9 d u = 9 u

\int \frac{13}{u} d u = 13 ln ∣ u ∣

= 9 u - 13 ln ∣ u ∣

Substitute back

u = t + 1

= 9 (t + 1) - 13 ln ∣ t + 1 ∣

Add a constant to the solution

= 9 (t + 1) - 13 ln ∣ t + 1 ∣ + C

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What is the integral of (9t-4)/(t+1) ?
The integral of (9t-4)/(t+1) is 9(t+1)-13ln|t+1|+C