What is the integral of ((50x+6))/((7x+1)(x-1)) ?

The integral of ((50x+6))/((7x+1)(x-1)) is 1/7 ln|7x+1|+7ln|x-1|+C

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integral of ((50x+6))/((7x+1)(x-1))

\int \frac{( 50 x + 6 )}{( 7 x + 1 ) ( x - 1 )} d x

\frac{1}{7} ln ∣ 7 x + 1 ∣ + 7 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{50 x + 6}{( 7 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{50 x + 6}{( 7 x + 1 ) ( x - 1 )} : \frac{1}{7 x + 1} + \frac{7}{x - 1}

= \int \frac{1}{7 x + 1} + \frac{7}{x - 1} d x

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{1}{7 x + 1} d x + \int \frac{7}{x - 1} d x

\int \frac{1}{7 x + 1} d x = \frac{1}{7} ln ∣ 7 x + 1 ∣

\int \frac{7}{x - 1} d x = 7 ln ∣ x - 1 ∣

= \frac{1}{7} ln ∣ 7 x + 1 ∣ + 7 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 7 x + 1 ∣ + 7 ln ∣ x - 1 ∣ + C

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

g (u) = u + 1 (1 - 8 u^{2})^{7}

\frac{d}{d x} (ln (1 + \frac{ln ( 2 )}{x}))

\int_{0}^{2} 2 x^{2} - x^{3} d x

\frac{d}{d x} (2 x^{2} + 4 x + 3)

What is the integral of ((50x+6))/((7x+1)(x-1)) ?
The integral of ((50x+6))/((7x+1)(x-1)) is 1/7 ln|7x+1|+7ln|x-1|+C