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

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

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integral of 1/(7x-3)

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

\frac{1}{7} ln ∣ 7 x - 3 ∣ + C

Solution steps

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

= \frac{1}{7} ln ∣ 7 x - 3 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 7 x - 3 ∣ + C

\int \frac{1}{3 x ^{5}} d x

x \to + (- 1) lim (- f (x))

f (x) = 4 x^{2} + 3 x

\frac{d}{dd} (\frac{e}{d})

x \to 0 + lim (3 x + 6)