What is the integral of 5/(2x+3) ?

The integral of 5/(2x+3) is 5/2 ln|2x+3|+C

Popular >

integral of 5/(2x+3)

\int \frac{5}{2 x + 3} d x

\frac{5}{2} ln ∣ 2 x + 3 ∣ + C

Solution steps

\int \frac{5}{2 x + 3} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{2 x + 3} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 3

= 5 \cdot \frac{1}{2} ln ∣ 2 x + 3 ∣

Simplify

5 \cdot \frac{1}{2} ln ∣ 2 x + 3 ∣ : \frac{5}{2} ln ∣ 2 x + 3 ∣

= \frac{5}{2} ln ∣ 2 x + 3 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x + 3 ∣ + C

f (x) = - x^{3} + 2 x^{2} - 2, at x = - 1

\frac{d y}{d x} + 2 y - 1 = 0

\frac{\partial}{\partial x} (x^{4} y - 4 x^{3} y^{2})

\int (\frac{3 x}{2} - \frac{9}{4}) e^{x (x - 3)} d x

f (x) = 1 - 2 tan (x), at x = \frac{3 π}{4}