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

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

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integral of 1/(2u+1)

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

\frac{1}{2} ln ∣ 2 u + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 v} d v

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

v = 2 u + 1

= \frac{1}{2} ln ∣ 2 u + 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 u + 1 ∣ + C

x \to 1 + lim (ln (3 x^{2} - x - 2))

f (x) = x^{2} + 1, at x = 3

x \to 3 lim (\frac{5}{( x - 3 ) ^{2}})

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y), y (1) = 0.5

\int x^{3} ln (x) d x