What is the integral of 5/(2x-2) ?

The integral of 5/(2x-2) is 5/2 ln|2x-2|+C

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integral of 5/(2x-2)

\int \frac{5}{2 x - 2} d x

\frac{5}{2} ln ∣ 2 x - 2 ∣ + C

Solution steps

\int \frac{5}{2 x - 2} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{2 x - 2} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 2

= 5 \cdot \frac{1}{2} ln ∣ 2 x - 2 ∣

Simplify

5 \cdot \frac{1}{2} ln ∣ 2 x - 2 ∣ : \frac{5}{2} ln ∣ 2 x - 2 ∣

= \frac{5}{2} ln ∣ 2 x - 2 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x - 2 ∣ + C

(1 + x^{6}) \frac{d y}{d x} = \frac{x ^{5}}{y}

\int_{0}^{4} \frac{e ^{x}}{x} d x

x \to 1 + lim (\frac{x}{ln ( x )})

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

y^{^{''}} - (\frac{x}{x - 1}) y^{^{'}} + \frac{1}{x - 1} y = 0