What is the integral of (5x+14)/((x+1)(x^2-4)) ?

The integral of (5x+14)/((x+1)(x^2-4)) is -3ln|x+1|+ln|x+2|+2ln|x-2|+C

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integral of (5x+14)/((x+1)(x^2-4))

\int \frac{5 x + 14}{( x + 1 ) ( x ^{2} - 4 )} d x

- 3 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{5 x + 14}{( x + 1 ) ( x ^{2} - 4 )} d x

Apply u-substitution

: \int \frac{5 u + 9}{u ( u ^{2} - 2 u - 3 )} d u

= \int \frac{5 u + 9}{u ( u ^{2} - 2 u - 3 )} d u

Take the partial fraction of

\frac{5 u + 9}{u ( u ^{2} - 2 u - 3 )} : - \frac{3}{u} + \frac{1}{u + 1} + \frac{2}{u - 3}

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

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

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{2}{u - 3} d u = 2 ln ∣ u - 3 ∣

= - 3 ln ∣ u ∣ + ln ∣ u + 1 ∣ + 2 ln ∣ u - 3 ∣

Substitute back

u = x + 1

= - 3 ln ∣ x + 1 ∣ + ln ∣ x + 1 + 1 ∣ + 2 ln ∣ x + 1 - 3 ∣

Simplify

= - 3 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - 3 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + 2 ln ∣ x - 2 ∣ + C

x \to \frac{π}{2} lim (e^{s e c (x)})

\frac{3}{( s + 2 ) ^{2}}

\int \frac{x ^{2} + x}{x ^{3} - x ^{2} - 2 x} d x

\int 5 (3 x - 1) 7 d x

ln (3 - x)

What is the integral of (5x+14)/((x+1)(x^2-4)) ?
The integral of (5x+14)/((x+1)(x^2-4)) is -3ln|x+1|+ln|x+2|+2ln|x-2|+C