What is the integral of 1/(3-7x) ?

The integral of 1/(3-7x) is -1/7 ln|3-7x|+C

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integral of 1/(3-7x)

\int \frac{1}{3 - 7 x} d x

- \frac{1}{7} ln ∣ 3 - 7 x ∣ + C

Solution steps

\int \frac{1}{3 - 7 x} d x

Apply u-substitution

: \int - \frac{1}{7 u} d u

= \int - \frac{1}{7 u} d u

Take the constant out:

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

= - \frac{1}{7} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{7} ln ∣ u ∣

Substitute back

u = 3 - 7 x

= - \frac{1}{7} ln ∣ 3 - 7 x ∣

Add a constant to the solution

= - \frac{1}{7} ln ∣ 3 - 7 x ∣ + C

f (x) = arcsin (x^{6})

\frac{\partial}{\partial x} (- \frac{x}{( x ^{2} + y ^{2} + 1 )})

\frac{d}{d x} (2 + ln (x))

\int \frac{361 - x ^{2}}{x} d x

\frac{d}{d x} (a cos (2 x))