What is the integral of (x+12)/(x^2+3x-10) ?

The integral of (x+12)/(x^2+3x-10) is -ln|x+5|+2ln|x-2|+C

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integral of (x+12)/(x^2+3x-10)

\int \frac{x + 12}{x ^{2} + 3 x - 10} d x

- ln ∣ x + 5 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x + 12}{x ^{2} + 3 x - 10} d x

Factor

x^{2} + 3 x - 10 : (x - 2) (x + 5)

= \int \frac{x + 12}{( x - 2 ) ( x + 5 )} d x

Apply u-substitution

: \int \frac{u + 14}{u ( u + 7 )} d u

= \int \frac{u + 14}{u ( u + 7 )} d u

Expand

\frac{u + 14}{u ( u + 7 )} : \frac{u}{u ( u + 7 )} + \frac{14}{u ( u + 7 )}

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{u}{u ( u + 7 )} d u + \int \frac{14}{u ( u + 7 )} d u

\int \frac{u}{u ( u + 7 )} d u = ln ∣ u + 7 ∣

\int \frac{14}{u ( u + 7 )} d u = 14 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

= ln ∣ u + 7 ∣ + 14 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = x - 2

= ln ∣ x - 2 + 7 ∣ + 14 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x - 2 + 7 ∣)

Simplify

ln ∣ x - 2 + 7 ∣ + 14 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x - 2 + 7 ∣) : - ln ∣ x + 5 ∣ + 2 ln ∣ x - 2 ∣

= - ln ∣ x + 5 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x + 5 ∣ + 2 ln ∣ x - 2 ∣ + C

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What is the integral of (x+12)/(x^2+3x-10) ?
The integral of (x+12)/(x^2+3x-10) is -ln|x+5|+2ln|x-2|+C