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

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

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

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

\frac{1}{2} ln ∣ 2 x + 3 ∣ + C

Solution steps

\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

\int x^{3} 1 - x^{2} d x

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

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{6}

\int \frac{1}{- x ^{2} - a + b} d x

x \to 1 lim (x^{\frac{1}{x ^{2} - 1}})

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