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

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

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

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

\frac{1}{8} ln ∣ x - 7 ∣ - \frac{1}{8} ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{8 ( x - 7 )} d x = \frac{1}{8} ln ∣ x - 7 ∣

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

= \frac{1}{8} ln ∣ x - 7 ∣ - \frac{1}{8} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x - 7 ∣ - \frac{1}{8} ln ∣ x + 1 ∣ + C

\frac{\partial}{\partial x} (6 + \frac{3}{x} + \frac{6}{x ^{2}})

(x + y e^{\frac{y}{x}}) d x = x e^{\frac{y}{x}} d y

\frac{d}{d x} (e^{3 l n (x^{2})})

x \to 4 lim (- 2 x)

\frac{2}{( s ^{2} + 4 s + 5 ) s}

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