What is the integral of 1/(4x+9) ?

The integral of 1/(4x+9) is 1/4 ln|4x+9|+C

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integral of 1/(4x+9)

\int \frac{1}{4 x + 9} d x

\frac{1}{4} ln ∣ 4 x + 9 ∣ + C

Solution steps

\int \frac{1}{4 x + 9} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x + 9

= \frac{1}{4} ln ∣ 4 x + 9 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x + 9 ∣ + C

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\int_{0}^{π} (π - x) sin (n x) d x

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

4 cos (4 t)

f (x) = (2 x^{3} + 3)^{6}