What is the integral of 5/(9+5x) ?

The integral of 5/(9+5x) is ln|9+5x|+C

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integral of 5/(9+5x)

\int \frac{5}{9 + 5 x} d x

ln ∣ 9 + 5 x ∣ + C

Solution steps

\int \frac{5}{9 + 5 x} 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}{9 + 5 x} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = 9 + 5 x

= 5 \cdot \frac{1}{5} ln ∣ 9 + 5 x ∣

Simplify

5 \cdot \frac{1}{5} ln ∣ 9 + 5 x ∣ : ln ∣ 9 + 5 x ∣

= ln ∣ 9 + 5 x ∣

Add a constant to the solution

= ln ∣ 9 + 5 x ∣ + C

\int_{0}^{a} x^{2} e^{x} d x

\frac{\partial}{\partial x} (6 x^{3} y^{2})

x \to 1 lim ((5 x - 1)^{- 2})

\frac{d}{d x} ((4 x - 4) (3 x^{3} - x^{2} + 1))

y = \frac{1}{16 x ^{2}}, y = x, y = \frac{x}{18}