What is the integral of 3/(5x-1) ?

The integral of 3/(5x-1) is 3/5 ln|5x-1|+C

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integral of 3/(5x-1)

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

\frac{3}{5} ln ∣ 5 x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{5 x - 1} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= 3 \cdot \int \frac{1}{5 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = 5 x - 1

= 3 \cdot \frac{1}{5} ln ∣ 5 x - 1 ∣

Simplify

3 \cdot \frac{1}{5} ln ∣ 5 x - 1 ∣ : \frac{3}{5} ln ∣ 5 x - 1 ∣

= \frac{3}{5} ln ∣ 5 x - 1 ∣

Add a constant to the solution

= \frac{3}{5} ln ∣ 5 x - 1 ∣ + C

\int_{0}^{2} (x 4 - x^{2}) d x

\frac{d}{d x} (2 + 3 x)

- 2 x^{2} + 8, 3 x + 1, 0

\int (x^{2} - 7 x + 4) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e^{- 2 x}