What is the integral of 1/(2x+9) ?

The integral of 1/(2x+9) is 1/2 ln|2x+9|+C

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integral of 1/(2x+9)

\int \frac{1}{2 x + 9} d x

\frac{1}{2} ln ∣ 2 x + 9 ∣ + C

Solution steps

\int \frac{1}{2 x + 9} 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 + 9

= \frac{1}{2} ln ∣ 2 x + 9 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 9 ∣ + C

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

y^{^{'}} = \frac{1 + x}{x y ^{12}}

n = 1 \sum \infty \frac{3}{n n}

f (x) = x^{3}, g (x) = 4 x

\frac{d}{d x} ((2 x^{3} + x^{2} - 6 x)^{\frac{2}{3}})