What is the integral of (1+2x)^{-1} ?

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

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integral of (1+2x)^{-1}

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

\frac{1}{2} ln ∣ 1 + 2 x ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

= \frac{1}{2} ln ∣ 1 + 2 x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 1 + 2 x ∣ + C

\int_{0}^{\infty} \frac{1}{x ^{2} + 4} d x

x^{4} e^{- x}

y = e^{2 e x}

\frac{\partial}{\partial x} (x^{3} - y^{3} + 6 x y)

\int 1 + e^{x} d x

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