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

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

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integral of 1/(2x-1)

integral

\frac{1}{2 x - 1}

\frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

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 x - 1

= \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

1 + t^{2}

\frac{( x ^{2} + 2 x + 1 )}{x ^{2}}

\frac{2}{( x ^{2} + 1 ) ^{2}}

\frac{1}{( x ^{2} + 2 ^{2} )}

2 + 2 cos^{4} (t)