What is the integral of 9/(6+9x) ?

The integral of 9/(6+9x) is ln|6+9x|+C

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integral of 9/(6+9x)

\int \frac{9}{6 + 9 x} d x

ln ∣ 6 + 9 x ∣ + C

Solution steps

\int \frac{9}{6 + 9 x} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{6 + 9 x} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= 9 \cdot \int \frac{1}{9 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{9} ln ∣ u ∣

Substitute back

u = 6 + 9 x

= 9 \cdot \frac{1}{9} ln ∣ 6 + 9 x ∣

Simplify

9 \cdot \frac{1}{9} ln ∣ 6 + 9 x ∣ : ln ∣ 6 + 9 x ∣

= ln ∣ 6 + 9 x ∣

Add a constant to the solution

= ln ∣ 6 + 9 x ∣ + C

\int (2^{x} + e^{π}) d x

e^{7}

d x + e^{3} x d y = 0

f (x) = x^{2}, (- 1, - 3)

x^{2} - 4 x, x