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

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

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integral of 1/(9+4x)

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

\frac{1}{4} ln ∣ 9 + 4 x ∣ + C

Solution steps

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

= \frac{1}{4} ln ∣ 9 + 4 x ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 9 + 4 x ∣ + C

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