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

Take the constant out:

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

= - \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

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

\int (sec^{2} (x) + x) d x

\int_{b}^{a} f (x) d x

\int \frac{x ^{3} + x + 1}{x ^{2} - x} d x

x \to 0 lim (e^{x^{2}})