What is the integral of (9t-5)/(t+1) ?

The integral of (9t-5)/(t+1) is 9(t+1)-14ln|t+1|+C

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integral of (9t-5)/(t+1)

\int \frac{9 t - 5}{t + 1} d t

9 (t + 1) - 14 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{9 t - 5}{t + 1} d t

Apply u-substitution

: \int \frac{9 u - 14}{u} d u

= \int \frac{9 u - 14}{u} d u

Expand

\frac{9 u - 14}{u} : 9 - \frac{14}{u}

= \int 9 - \frac{14}{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{14}{u} d u

\int 9 d u = 9 u

\int \frac{14}{u} d u = 14 ln ∣ u ∣

= 9 u - 14 ln ∣ u ∣

Substitute back

u = t + 1

= 9 (t + 1) - 14 ln ∣ t + 1 ∣

Add a constant to the solution

= 9 (t + 1) - 14 ln ∣ t + 1 ∣ + C

x \to 0 lim (\frac{x - 32}{x - 4})

\int 7 x - 14 d x

x \to - 1 - lim (x^{2} + 2)

f (x) = \frac{x ^{2} + 2 x}{x ^{2} - 1}

\int \frac{1}{1 + 14 x} d x

What is the integral of (9t-5)/(t+1) ?
The integral of (9t-5)/(t+1) is 9(t+1)-14ln|t+1|+C