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

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

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integral of 2/(x-1)

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

2 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = x - 1

= 2 ln ∣ x - 1 ∣

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + C

\int \frac{4 x + 6}{x ^{2} + 3 x + 1} d x

y = \frac{( 2 x ^{2} + 7 )}{x ^{2} + 3}

x \to 3 π + lim (csc (x))

f (x) = - \frac{5}{x ^{2}}

4 x y (\frac{d y}{d x}) = 1