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

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

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integral of 1/(1+2x)

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

\frac{1}{2} ln ∣ 1 + 2 x ∣ + C

Solution steps

\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

x \to 3 lim (4 x)

x \to 3 lim (5 x^{3} - 4 x^{2} + x - 6)

\frac{d}{d x} ((a x^{2} + b x + c) e^{- \frac{x}{2}})

\frac{d ^{4}}{d x ^{4}} (e^{\frac{6}{x}})

f (x) = ln (16 sin^{2} (x))