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

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

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integral of 1/(2u)

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

\frac{1}{2} ln ∣ u ∣ + C

Solution steps

\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 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ u ∣ + C

121 y^{^{''}} + 198 y^{^{'}} + 65 y = 0

(x^{2} - 6 x + 8)^{4}

(1 + ln (x) + \frac{y}{x}) d x = (1 - ln (x)) d y

\frac{d}{d x} (7 - 4 e^{- x})

(- \frac{q ^{4}}{4} + \frac{2 q ^{3}}{3} + \frac{11 q ^{2}}{2} - 12 q)^{^{'}}