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

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

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

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

x + \frac{3}{2} - 2 ln ∣ 2 x + 3 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 4}{2 u} d u

= \int \frac{u - 4}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int 1 - \frac{4}{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

= \frac{1}{2} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{2} (u - 4 ln ∣ u ∣)

Substitute back

u = 2 x + 3

= \frac{1}{2} (2 x + 3 - 4 ln ∣ 2 x + 3 ∣)

Simplify

\frac{1}{2} (2 x + 3 - 4 ln ∣ 2 x + 3 ∣) : x + \frac{3}{2} - 2 ln ∣ 2 x + 3 ∣

= x + \frac{3}{2} - 2 ln ∣ 2 x + 3 ∣

Add a constant to the solution

= x + \frac{3}{2} - 2 ln ∣ 2 x + 3 ∣ + C

\frac{d}{d x} (ln (x + 2 y + 3 z))

y^{^{'''}} = e^{- 0.2 x}

\frac{( s + 2 )}{( s ^{2} + 4 )}

(x^{2} + 4) (2 x - 6)

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

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