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

\int \frac{3 e ^{- x} + 2 e ^{3 x}}{e ^{2 x}} d x

\frac{d}{d x} (in (x))

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

2 x^{3} - x^{2}, 0, 1

\frac{d y}{d t} - 2 y = y^{4}

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