What is the integral of ((3t-2))/(t+1) ?

The integral of ((3t-2))/(t+1) is 3(t+1)-5ln|t+1|+C

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integral of ((3t-2))/(t+1)

\int \frac{( 3 t - 2 )}{t + 1} d t

3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{3 t - 2}{t + 1} d t

Apply u-substitution

: \int \frac{3 u - 5}{u} d u

= \int \frac{3 u - 5}{u} d u

Expand

\frac{3 u - 5}{u} : 3 - \frac{5}{u}

= \int 3 - \frac{5}{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 3 d u - \int \frac{5}{u} d u

\int 3 d u = 3 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= 3 u - 5 ln ∣ u ∣

Substitute back

u = t + 1

= 3 (t + 1) - 5 ln ∣ t + 1 ∣

Add a constant to the solution

= 3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

\frac{3}{x ^{5}}

y^{^{'}} + 5 y = 15

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

\int x (cos (x) - sin (x)) d x

\frac{d}{d x} (x - 0.8 - 0.2 \cdot sin (x))

What is the integral of ((3t-2))/(t+1) ?
The integral of ((3t-2))/(t+1) is 3(t+1)-5ln|t+1|+C