What is the integral of 4/(x-1) ?

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

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integral of 4/(x-1)

\int \frac{4}{x - 1} d x

4 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{4}{x - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x - 1} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 ln ∣ u ∣

Substitute back

u = x - 1

= 4 ln ∣ x - 1 ∣

Add a constant to the solution

= 4 ln ∣ x - 1 ∣ + C

\frac{6 s - 4}{s ^{2} - 4 s + 20}

x \to 0 lim (x cos (\frac{1}{x ^{3}}))

y^{^{'}} + x y = x^{3}

y = 2 x, y = \frac{x ^{2}}{4}

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