What is the integral of 7/((x-3)(x-5)) ?

The integral of 7/((x-3)(x-5)) is 7(-1/2 ln|x-3|+1/2 ln|x-5|)+C

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integral of 7/((x-3)(x-5))

\int \frac{7}{( x - 3 ) ( x - 5 )} d x

7 (- \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{7}{( x - 3 ) ( x - 5 )} 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 - 3 ) ( x - 5 )} d x

Take the partial fraction of

\frac{1}{( x - 3 ) ( x - 5 )} : - \frac{1}{2 ( x - 3 )} + \frac{1}{2 ( x - 5 )}

= 7 \cdot \int - \frac{1}{2 ( x - 3 )} + \frac{1}{2 ( 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

= 7 (- \int \frac{1}{2 ( x - 3 )} d x + \int \frac{1}{2 ( x - 5 )} d x)

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

\int \frac{1}{2 ( x - 5 )} d x = \frac{1}{2} ln ∣ x - 5 ∣

= 7 (- \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣)

Add a constant to the solution

= 7 (- \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣) + C

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What is the integral of 7/((x-3)(x-5)) ?
The integral of 7/((x-3)(x-5)) is 7(-1/2 ln|x-3|+1/2 ln|x-5|)+C