What is the integral of ((65x+7))/((8x+1)(x-1)) ?

The integral of ((65x+7))/((8x+1)(x-1)) is 1/8 ln|8x+1|+8ln|x-1|+C

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integral of ((65x+7))/((8x+1)(x-1))

\int \frac{( 65 x + 7 )}{( 8 x + 1 ) ( x - 1 )} d x

\frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{65 x + 7}{( 8 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{65 x + 7}{( 8 x + 1 ) ( x - 1 )} : \frac{1}{8 x + 1} + \frac{8}{x - 1}

= \int \frac{1}{8 x + 1} + \frac{8}{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}{8 x + 1} d x + \int \frac{8}{x - 1} d x

\int \frac{1}{8 x + 1} d x = \frac{1}{8} ln ∣ 8 x + 1 ∣

\int \frac{8}{x - 1} d x = 8 ln ∣ x - 1 ∣

= \frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣ + C

\int \frac{10}{e ^{\frac{x}{2}}} d x

\int \frac{tan ^{3} ( x )}{sec ( x )} d x

\frac{\partial}{\partial y} (x^{4} y - x^{2} y^{3})

\frac{\partial}{\partial x} (1 - (x^{2} + y^{2})^{\frac{1}{7}})

x \to 3 - lim (e^{\frac{7}{( x - 3 )}})

What is the integral of ((65x+7))/((8x+1)(x-1)) ?
The integral of ((65x+7))/((8x+1)(x-1)) is 1/8 ln|8x+1|+8ln|x-1|+C