What is the integral of (x+4)/(x^2+9x-10) ?

The integral of (x+4)/(x^2+9x-10) is ln|x+10|+5(1/11 ln|x-1|-1/11 ln|x+10|)+C

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integral of (x+4)/(x^2+9x-10)

\int \frac{x + 4}{x ^{2} + 9 x - 10} d x

ln ∣ x + 10 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x + 10 ∣) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 9 x - 10} d x

Factor

x^{2} + 9 x - 10 : (x - 1) (x + 10)

= \int \frac{x + 4}{( x - 1 ) ( x + 10 )} d x

Apply u-substitution

= \int \frac{u + 5}{u ( u + 11 )} d u

Expand

\frac{u + 5}{u ( u + 11 )} : \frac{u}{u ( u + 11 )} + \frac{5}{u ( u + 11 )}

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 + 11 )} d u + \int \frac{5}{u ( u + 11 )} d u

\int \frac{u}{u ( u + 11 )} d u = ln ∣ u + 11 ∣

\int \frac{5}{u ( u + 11 )} d u = 5 (\frac{1}{11} ln ∣ u ∣ - \frac{1}{11} ln ∣ u + 11 ∣)

= ln ∣ u + 11 ∣ + 5 (\frac{1}{11} ln ∣ u ∣ - \frac{1}{11} ln ∣ u + 11 ∣)

Substitute back

u = x - 1

= ln ∣ x - 1 + 11 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x - 1 + 11 ∣)

Simplify

ln ∣ x - 1 + 11 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x - 1 + 11 ∣) : ln ∣ x + 10 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x + 10 ∣)

= ln ∣ x + 10 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x + 10 ∣)

Add a constant to the solution

= ln ∣ x + 10 ∣ + 5 (\frac{1}{11} ln ∣ x - 1 ∣ - \frac{1}{11} ln ∣ x + 10 ∣) + C

What is the integral of (x+4)/(x^2+9x-10) ?
The integral of (x+4)/(x^2+9x-10) is ln|x+10|+5(1/11 ln|x-1|-1/11 ln|x+10|)+C