What is the integral of (9t-7)/(t+1) ?

The integral of (9t-7)/(t+1) is 9(t+1)-16ln|t+1|+C

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integral of (9t-7)/(t+1)

\int \frac{9 t - 7}{t + 1} d t

9 (t + 1) - 16 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{9 t - 7}{t + 1} d t

Apply u-substitution

: \int \frac{9 u - 16}{u} d u

= \int \frac{9 u - 16}{u} d u

Expand

\frac{9 u - 16}{u} : 9 - \frac{16}{u}

= \int 9 - \frac{16}{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{16}{u} d u

\int 9 d u = 9 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= 9 u - 16 ln ∣ u ∣

Substitute back

u = t + 1

= 9 (t + 1) - 16 ln ∣ t + 1 ∣

Add a constant to the solution

= 9 (t + 1) - 16 ln ∣ t + 1 ∣ + C

x \to - 3 lim (\frac{x - 3}{x ^{2} - 9})

\int e^{2 x} (1 + e^{2 x})^{3} d x

\frac{d}{d w} (2 l + 2 w)

\int (t^{2} + 1) d t

\frac{\partial}{\partial x} (2 x^{3} + 2 x^{2} y - 3 y^{3})

What is the integral of (9t-7)/(t+1) ?
The integral of (9t-7)/(t+1) is 9(t+1)-16ln|t+1|+C