What is the integral of 1/(9x+7) ?

The integral of 1/(9x+7) is 1/9 ln|9x+7|+C

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integral of 1/(9x+7)

\int \frac{1}{9 x + 7} d x

\frac{1}{9} ln ∣ 9 x + 7 ∣ + C

Solution steps

\int \frac{1}{9 x + 7} d x

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 x + 7

= \frac{1}{9} ln ∣ 9 x + 7 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x + 7 ∣ + C

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

\int (\frac{9}{x} - \frac{4}{x ^{3}}) d x

\frac{d}{d θ} (cos (θ) - sin (θ))

im pl i c i t \frac{d y}{d x}, y^{3} = x^{4}

\frac{\partial}{\partial x} (\frac{x}{y} + e^{y})