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

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

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integral of (3x+2)/(x+1)

\int \frac{3 x + 2}{x + 1} d x

3 (x + 1) - ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{3 x + 2}{x + 1} d x

Apply u-substitution

: \int \frac{3 u - 1}{u} d u

= \int \frac{3 u - 1}{u} d u

Expand

\frac{3 u - 1}{u} : 3 - \frac{1}{u}

= \int 3 - \frac{1}{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{1}{u} d u

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 3 u - ln ∣ u ∣

Substitute back

u = x + 1

= 3 (x + 1) - ln ∣ x + 1 ∣

Add a constant to the solution

= 3 (x + 1) - ln ∣ x + 1 ∣ + C

2 x^{3} + 10 x^{2} + 22 x - 3, x^{3} - x^{2} - 6 x - 3, - 2, - 2

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

y = x^{2} + x, (1, 2)

f (x) = - 3 cos (x), at x = \frac{π}{2}

x^{2} + 12 x, 3 x^{2} + 10, 1, 4

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