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))

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

\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

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(sin (x) cos (y)) d x + (cos (x) sin (y)) d y = 0

- x^{4} - 6 x^{3} + 8 x - 6

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What is the integral of 1/((2x+1)) ?
The integral of 1/((2x+1)) is 1/2 ln|2x+1|+C