What is the integral of (2z-3)/(7z-10) ?

The integral of (2z-3)/(7z-10) is 2/7 z-20/49-1/49 ln|7z-10|+C

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integral of (2z-3)/(7z-10)

\int \frac{2 z - 3}{7 z - 10} d z

\frac{2}{7} z - \frac{20}{49} - \frac{1}{49} ln ∣ 7 z - 10 ∣ + C

Solution steps

\int \frac{2 z - 3}{7 z - 10} d z

Expand

\frac{2 z - 3}{7 z - 10} : \frac{2 z}{7 z - 10} - \frac{3}{7 z - 10}

Apply the Sum Rule:

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

= \int \frac{2 z}{7 z - 10} d z - \int \frac{3}{7 z - 10} d z

\int \frac{2 z}{7 z - 10} d z = \frac{2}{49} (7 z - 10 + 10 ln ∣ 7 z - 10 ∣)

\int \frac{3}{7 z - 10} d z = \frac{3}{7} ln ∣ 7 z - 10 ∣

= \frac{2}{49} (7 z - 10 + 10 ln ∣ 7 z - 10 ∣) - \frac{3}{7} ln ∣ 7 z - 10 ∣

Simplify

\frac{2}{49} (7 z - 10 + 10 ln ∣ 7 z - 10 ∣) - \frac{3}{7} ln ∣ 7 z - 10 ∣ : \frac{2}{7} z - \frac{20}{49} - \frac{1}{49} ln ∣ 7 z - 10 ∣

= \frac{2}{7} z - \frac{20}{49} - \frac{1}{49} ln ∣ 7 z - 10 ∣

Add a constant to the solution

= \frac{2}{7} z - \frac{20}{49} - \frac{1}{49} ln ∣ 7 z - 10 ∣ + C

\frac{\partial}{\partial x} (2 sin (x) cos (y))

\frac{d}{d x} (\frac{x}{x ^{- 1} + 1})

\frac{\partial}{\partial x} (x^{3} y + 12 x^{2} - 64 y)

\int (3 + 3 tan^{2} (x)) d x

\frac{5 x ^{3}}{( 5 - 3 x ) ^{4}}

What is the integral of (2z-3)/(7z-10) ?
The integral of (2z-3)/(7z-10) is 2/7 z-20/49-1/49 ln|7z-10|+C