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

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

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

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

ln ∣ x - 1 ∣ - ln ∣ x + 6 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 7 \cdot \int \frac{1}{7 ( x - 1 )} - \frac{1}{7 ( x + 6 )} 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

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

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

\int \frac{1}{7 ( x + 6 )} d x = \frac{1}{7} ln ∣ x + 6 ∣

= 7 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣)

Simplify

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

= ln ∣ x - 1 ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ - ln ∣ x + 6 ∣ + C

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