What is the integral of 1/(2-2x) ?

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

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integral of 1/(2-2x)

\int \frac{1}{2 - 2 x} d x

- \frac{1}{2} ln ∣ 2 - 2 x ∣ + C

Solution steps

\int \frac{1}{2 - 2 x} d x

Apply u-substitution

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

= \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 ∣

Substitute back

u = 2 - 2 x

= - \frac{1}{2} ln ∣ 2 - 2 x ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 2 - 2 x ∣ + C

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\frac{d}{d x} (x^{2} - 4 x)

(1 + x^{2}) d y + (x y + x^{3} + x) d x = 0

\frac{d y}{d x} = \frac{( 6 x - 2 x y ^{3} )}{y ^{2}}

2 x^{2} + 3 x - 2