What is the integral of (5x)/((2x+1)(x-7)) ?

The integral of (5x)/((2x+1)(x-7)) is 1/6 (ln|2x+1|+14ln|2x-14|)+C

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integral of (5x)/((2x+1)(x-7))

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

\frac{1}{6} (ln ∣ 2 x + 1 ∣ + 14 ln ∣ 2 x - 14 ∣) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{2 u ( u - 15 )} d u

= 5 \cdot \int \frac{u - 1}{2 u ( u - 15 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{u - 1}{u ( u - 15 )} d u

Expand

\frac{u - 1}{u ( u - 15 )} : \frac{u}{u ( u - 15 )} - \frac{1}{u ( u - 15 )}

Apply the Sum Rule:

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

= 5 \cdot \frac{1}{2} (\int \frac{u}{u ( u - 15 )} d u - \int \frac{1}{u ( u - 15 )} d u)

\int \frac{u}{u ( u - 15 )} d u = ln ∣ u - 15 ∣

\int \frac{1}{u ( u - 15 )} d u = - \frac{1}{15} ln ∣ u ∣ + \frac{1}{15} ln ∣ u - 15 ∣

= 5 \cdot \frac{1}{2} (ln ∣ u - 15 ∣ - (- \frac{1}{15} ln ∣ u ∣ + \frac{1}{15} ln ∣ u - 15 ∣))

Substitute back

u = 2 x + 1

= 5 \cdot \frac{1}{2} (ln ∣ 2 x + 1 - 15 ∣ - (- \frac{1}{15} ln ∣ 2 x + 1 ∣ + \frac{1}{15} ln ∣ 2 x + 1 - 15 ∣))

Simplify

5 \cdot \frac{1}{2} (ln ∣ 2 x + 1 - 15 ∣ - (- \frac{1}{15} ln ∣ 2 x + 1 ∣ + \frac{1}{15} ln ∣ 2 x + 1 - 15 ∣)) : \frac{1}{6} (ln ∣ 2 x + 1 ∣ + 14 ln ∣ 2 x - 14 ∣)

= \frac{1}{6} (ln ∣ 2 x + 1 ∣ + 14 ln ∣ 2 x - 14 ∣)

Add a constant to the solution

= \frac{1}{6} (ln ∣ 2 x + 1 ∣ + 14 ln ∣ 2 x - 14 ∣) + C

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What is the integral of (5x)/((2x+1)(x-7)) ?
The integral of (5x)/((2x+1)(x-7)) is 1/6 (ln|2x+1|+14ln|2x-14|)+C