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

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

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integral of 2/(2x+1)

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

ln ∣ 2 x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{2 x + 1} d x

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 1

= 2 \cdot \frac{1}{2} ln ∣ 2 x + 1 ∣

Simplify

2 \cdot \frac{1}{2} ln ∣ 2 x + 1 ∣ : ln ∣ 2 x + 1 ∣

= ln ∣ 2 x + 1 ∣

Add a constant to the solution

= ln ∣ 2 x + 1 ∣ + C

\int (e^{x})^{2} d x

n = 1 \sum \infty 23 (\frac{2}{5})^{n}

\int \frac{e ^{x}}{3 - e ^{2 x}} d x

y (x + 3) + y^{^{'}} = 0

8 + 2 x + x e^{x}