What is the integral of 4/((x-9)(x-5)) ?

The integral of 4/((x-9)(x-5)) is ln|x-9|-ln|x-5|+C

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integral of 4/((x-9)(x-5))

\int \frac{4}{( x - 9 ) ( x - 5 )} d x

ln ∣ x - 9 ∣ - ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{4}{( x - 9 ) ( x - 5 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x - 9 ) ( x - 5 )} d x

Take the partial fraction of

\frac{1}{( x - 9 ) ( x - 5 )} : \frac{1}{4 ( x - 9 )} - \frac{1}{4 ( x - 5 )}

= 4 \cdot \int \frac{1}{4 ( x - 9 )} - \frac{1}{4 ( x - 5 )} 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

= 4 (\int \frac{1}{4 ( x - 9 )} d x - \int \frac{1}{4 ( x - 5 )} d x)

\int \frac{1}{4 ( x - 9 )} d x = \frac{1}{4} ln ∣ x - 9 ∣

\int \frac{1}{4 ( x - 5 )} d x = \frac{1}{4} ln ∣ x - 5 ∣

= 4 (\frac{1}{4} ln ∣ x - 9 ∣ - \frac{1}{4} ln ∣ x - 5 ∣)

Simplify

4 (\frac{1}{4} ln ∣ x - 9 ∣ - \frac{1}{4} ln ∣ x - 5 ∣) : ln ∣ x - 9 ∣ - ln ∣ x - 5 ∣

= ln ∣ x - 9 ∣ - ln ∣ x - 5 ∣

Add a constant to the solution

= ln ∣ x - 9 ∣ - ln ∣ x - 5 ∣ + C

\frac{5}{x}

\int 5 x e^{8 x^{2}} d x

\int e^{x^{7}} x^{6} d x

\int \frac{3 e ^{3 x}}{x} d x

\int \frac{3 + 9 x}{1 + x ^{2}} d x

What is the integral of 4/((x-9)(x-5)) ?
The integral of 4/((x-9)(x-5)) is ln|x-9|-ln|x-5|+C