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

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

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integral of u/(2u+1)

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

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

Solution steps

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

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

Add a constant to the solution

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

x^{3}, x + 6 = y, \frac{- x}{2} = y

\frac{\partial}{\partial y} (ln (x y))

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

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{- x}

\int - csc (2 w) cot (2 w) d w

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