What is the integral of (2x+3)^{-1} ?

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

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integral of (2x+3)^{-1}

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

\frac{1}{2} ln ∣ 2 x + 3 ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 3

= \frac{1}{2} ln ∣ 2 x + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 3 ∣ + C

\frac{d}{d s} (\frac{s - s}{s ^{2}})

3 - 2 x

\frac{d}{d x} ((2 x + 1)^{3} e^{- (1 - x)^{4}})

f (x) = \frac{1}{x + 6}

\frac{d}{d v} (\frac{1}{v})

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