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

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

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integral of 1/(1-7x)

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

- \frac{1}{7} ln ∣ 1 - 7 x ∣ + C

Solution steps

\int \frac{1}{1 - 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 = 1 - 7 x

= - \frac{1}{7} ln ∣ 1 - 7 x ∣

Add a constant to the solution

= - \frac{1}{7} ln ∣ 1 - 7 x ∣ + C

y^{^{''}} - 2 y^{^{'}} + y = - 2 cos (x) - 2

n = 0 \sum \infty \frac{n}{2 n}

h \to 0 lim (\frac{h}{sin ( 3 h )})

\frac{\partial}{\partial x} (4 x^{2} - 2 y)

\frac{d}{d x} (ln ((sin (x) cos (x))^{2}))