What is the integral of (7x-23)/(2(x-2)(x-5)) ?

The integral of (7x-23)/(2(x-2)(x-5)) is 1/2 (3ln|x-2|+4ln|x-5|)+C

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integral of (7x-23)/(2(x-2)(x-5))

\int \frac{7 x - 23}{2 ( x - 2 ) ( x - 5 )} d x

\frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{7 x - 23}{2 ( x - 2 ) ( x - 5 )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{7 x - 23}{( x - 2 ) ( x - 5 )} d x

Apply u-substitution

: \int \frac{7 u - 9}{u ( u - 3 )} d u

= \frac{1}{2} \cdot \int \frac{7 u - 9}{u ( u - 3 )} d u

Take the partial fraction of

\frac{7 u - 9}{u ( u - 3 )} : \frac{3}{u} + \frac{4}{u - 3}

= \frac{1}{2} \cdot \int \frac{3}{u} + \frac{4}{u - 3} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{3}{u} d u + \int \frac{4}{u - 3} d u)

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int \frac{4}{u - 3} d u = 4 ln ∣ u - 3 ∣

= \frac{1}{2} (3 ln ∣ u ∣ + 4 ln ∣ u - 3 ∣)

Substitute back

u = x - 2

= \frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 2 - 3 ∣)

Simplify

\frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 2 - 3 ∣) : \frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 5 ∣)

= \frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 5 ∣)

Add a constant to the solution

= \frac{1}{2} (3 ln ∣ x - 2 ∣ + 4 ln ∣ x - 5 ∣) + C

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What is the integral of (7x-23)/(2(x-2)(x-5)) ?
The integral of (7x-23)/(2(x-2)(x-5)) is 1/2 (3ln|x-2|+4ln|x-5|)+C