What is the integral of (x+4)/(x^2+7x-8) ?

The integral of (x+4)/(x^2+7x-8) is ln|x+8|+5(1/9 ln|x-1|-1/9 ln|x+8|)+C

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integral of (x+4)/(x^2+7x-8)

\int \frac{x + 4}{x ^{2} + 7 x - 8} d x

ln ∣ x + 8 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 7 x - 8} d x

Factor

x^{2} + 7 x - 8 : (x - 1) (x + 8)

= \int \frac{x + 4}{( x - 1 ) ( x + 8 )} d x

Apply u-substitution

: \int \frac{u + 5}{u ( u + 9 )} d u

= \int \frac{u + 5}{u ( u + 9 )} d u

Expand

\frac{u + 5}{u ( u + 9 )} : \frac{u}{u ( u + 9 )} + \frac{5}{u ( u + 9 )}

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 + 9 )} d u + \int \frac{5}{u ( u + 9 )} d u

\int \frac{u}{u ( u + 9 )} d u = ln ∣ u + 9 ∣

\int \frac{5}{u ( u + 9 )} d u = 5 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

= ln ∣ u + 9 ∣ + 5 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

Substitute back

u = x - 1

= ln ∣ x - 1 + 9 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x - 1 + 9 ∣)

Simplify

ln ∣ x - 1 + 9 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x - 1 + 9 ∣) : ln ∣ x + 8 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣)

= ln ∣ x + 8 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣)

Add a constant to the solution

= ln ∣ x + 8 ∣ + 5 (\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 8 ∣) + C

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What is the integral of (x+4)/(x^2+7x-8) ?
The integral of (x+4)/(x^2+7x-8) is ln|x+8|+5(1/9 ln|x-1|-1/9 ln|x+8|)+C