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

The integral of u/(1+2u) is 1/4 (1+2u-ln|1+2u|)+C

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integral of u/(1+2u)

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

\frac{1}{4} (1 + 2 u - ln ∣ 1 + 2 u ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{v - 1}{4 v} d v

= \int \frac{v - 1}{4 v} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{4} \cdot \int 1 - \frac{1}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{4} (v - ln ∣ v ∣)

Substitute back

v = 1 + 2 u

= \frac{1}{4} (1 + 2 u - ln ∣ 1 + 2 u ∣)

Add a constant to the solution

= \frac{1}{4} (1 + 2 u - ln ∣ 1 + 2 u ∣) + C

f (x) = 2 x^{2} - 12 x + 8

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

y = (1 + sin (7 t))^{- 6}

\frac{d y}{d t} = \frac{t e ^{t}}{y 9 + y ^{2}}

What is the integral of u/(1+2u) ?
The integral of u/(1+2u) is 1/4 (1+2u-ln|1+2u|)+C