What is the integral of (6-y)/(y-1) ?

The integral of (6-y)/(y-1) is -y+1+5ln|y-1|+C

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integral of (6-y)/(y-1)

\int \frac{6 - y}{y - 1} d y

- y + 1 + 5 ln ∣ y - 1 ∣ + C

Solution steps

\int \frac{6 - y}{y - 1} d y

Apply u-substitution

: \int \frac{- u + 5}{u} d u

= \int \frac{- u + 5}{u} d u

Expand

\frac{- u + 5}{u} : - 1 + \frac{5}{u}

= \int - 1 + \frac{5}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int \frac{5}{u} d u

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= - u + 5 ln ∣ u ∣

Substitute back

u = y - 1

= - (y - 1) + 5 ln ∣ y - 1 ∣

Simplify

= - y + 1 + 5 ln ∣ y - 1 ∣

Add a constant to the solution

= - y + 1 + 5 ln ∣ y - 1 ∣ + C

\int_{2}^{8} (10 - x) d x

f (x) = (2 x^{7} - 3 x^{5} + 2)^{22}

\frac{s + 3}{s ^{2} + 3 s + 2}

\frac{\partial}{\partial x} (x y - z)

f (x) = \frac{x ^{2} + 6}{9 x - 2}

What is the integral of (6-y)/(y-1) ?
The integral of (6-y)/(y-1) is -y+1+5ln|y-1|+C