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

y = \frac{- 12}{4 x}

2 x + y + (x + 2 y) y^{^{'}} = 0

\int - 2 x^{2} + 6 x d x

(cos^{2} (s) - 11 s^{2})^{2}

\frac{d}{d x} (- 3 cot (x) csc (x))