What is the integral of 9/((x-1)(x+8)) ?

The integral of 9/((x-1)(x+8)) is ln|x-1|-ln|x+8|+C

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integral of 9/((x-1)(x+8))

\int \frac{9}{( x - 1 ) ( x + 8 )} d x

ln ∣ x - 1 ∣ - ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{9}{( x - 1 ) ( x + 8 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( x - 1 ) ( x + 8 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x + 8 )} : \frac{1}{9 ( x - 1 )} - \frac{1}{9 ( x + 8 )}

= 9 \cdot \int \frac{1}{9 ( x - 1 )} - \frac{1}{9 ( x + 8 )} 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

= 9 (\int \frac{1}{9 ( x - 1 )} d x - \int \frac{1}{9 ( x + 8 )} d x)

\int \frac{1}{9 ( x - 1 )} d x = \frac{1}{9} ln ∣ x - 1 ∣

\int \frac{1}{9 ( x + 8 )} d x = \frac{1}{9} ln ∣ x + 8 ∣

= 9 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣)

Simplify

9 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣) : ln ∣ x - 1 ∣ - ln ∣ x + 8 ∣

= ln ∣ x - 1 ∣ - ln ∣ x + 8 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ - ln ∣ x + 8 ∣ + C

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 1

\frac{d}{d t} (6 ln (cos (t)))

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

\int 5 e^{- 5 x} d x

\frac{d}{d x} (arctan (\frac{1}{1 + x ^{2}}))

What is the integral of 9/((x-1)(x+8)) ?
The integral of 9/((x-1)(x+8)) is ln|x-1|-ln|x+8|+C