What is the integral of 1/(2y-1) ?

The integral of 1/(2y-1) is 1/2 ln|2y-1|+C

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integral of 1/(2y-1)

\int \frac{1}{2 y - 1} d y

\frac{1}{2} ln ∣ 2 y - 1 ∣ + C

Solution steps

\int \frac{1}{2 y - 1} d y

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 y - 1

= \frac{1}{2} ln ∣ 2 y - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 y - 1 ∣ + C

y = (x^{3} + 2) (3 x^{2} - 4 x + 2), (1, 3)

x^{2} (2.4)

\int 7 x + 5 d x

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