What is the integral of (9t-8)/(t+1) ?

The integral of (9t-8)/(t+1) is 9(t+1)-17ln|t+1|+C

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integral of (9t-8)/(t+1)

\int \frac{9 t - 8}{t + 1} d t

9 (t + 1) - 17 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{9 t - 8}{t + 1} d t

Apply u-substitution

: \int \frac{9 u - 17}{u} d u

= \int \frac{9 u - 17}{u} d u

Expand

\frac{9 u - 17}{u} : 9 - \frac{17}{u}

= \int 9 - \frac{17}{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{17}{u} d u

\int 9 d u = 9 u

\int \frac{17}{u} d u = 17 ln ∣ u ∣

= 9 u - 17 ln ∣ u ∣

Substitute back

u = t + 1

= 9 (t + 1) - 17 ln ∣ t + 1 ∣

Add a constant to the solution

= 9 (t + 1) - 17 ln ∣ t + 1 ∣ + C

\frac{\partial}{\partial x} (x y (6 - x - y))

\frac{d}{d x} (x^{\frac{4}{3}} - 4 x^{\frac{1}{3}} + 1)

\frac{\partial}{\partial x} (cos (4 y - x))

\frac{\partial}{\partial x} (4 y^{x})

9 x - 9 x

What is the integral of (9t-8)/(t+1) ?
The integral of (9t-8)/(t+1) is 9(t+1)-17ln|t+1|+C